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The vertex of the pyramid and the center of its square base both lie on the same diameter of the sphere. The pyramid has a vertical height \$h\$ and the side length of its square base is \$s\$, such that \$0 < h < 2K\$.","difficulty":7,"difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"3f2a2356-b41c-4950-8022-379e25ef2de4","label":"a","content":"Show that the volume, \$V\$, of the pyramid is given by\n\\[V = \\frac{2}{3}(2Kh^2 - h^3)\\]","markscheme":"\n- Identifies that the distance from the center of the sphere to a vertex of the square base is $\\frac{s}{\\sqrt{2}}$ and the distance from the sphere center to the pyramid base is $|h - K|$ M1\n- Applies Pythagoras' Theorem to the right-angled triangle formed by the sphere radius, the base half-diagonal, and the center-to-base distance: $(\\frac{s}{\\sqrt{2}})^2 + (h - K)^2 = K^2$ A1\n- Expands and rearranges to find $s^2 = 4Kh - 2h^2$, then substitutes into $V = \\frac{1}{3}s^2h$ to reach $V = \\frac{2}{3}(2Kh^2 - h^3)$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"pythagorean","label":"Pythagorean Theorem","steps":[{"type":"text","order":0,"title":"Recall the volume formula","content":"To find the volume $V$ of a right square-based pyramid, we use the standard formula:\n\n$$V = \\frac{1}{3} \\times \\text{base area} \\times \\text{height}$$\n\nSince the base of the pyramid is a square with side length $s$, its area is $s^2$. The vertical height is given as $h$. Substituting these into the formula yields:\n\n$$V = \\frac{1}{3}s^2h$$\n\nTo show that $V = \\frac{2}{3}(2Kh^2 - h^3)$, we need to express $s^2$ in terms of the sphere's radius $K$ and the pyramid's height $h$.","highlights":[{"text":"Show that the volume, \$V\$, of the pyramid is given by\n\\[V = \\frac{2}{3}(2Kh^2 - h^3)\\]","location":"specification"}]},{"type":"text","order":1,"title":"Define the right-angled triangle","content":"Let the center of the solid crystal sphere be $O$, the center of the pyramid's square base be $C$, and one of the vertices of the square base be $P$.\n\nWe can form a right-angled triangle $\\triangle OCP$:\n- The hypotenuse $OP$ is the radius of the sphere, so $OP = K$.\n- The side $OC$ is the vertical distance from the center of the sphere to the base of the pyramid. Since the total height of the pyramid is $h$ from the apex (which lies on the sphere's surface) to the base, and the radius from the apex to the sphere's center is $K$, this distance is given by:\n $$OC = |h - K|$$\n- The side $CP$ is the distance from the center of the square base to a corner vertex. For a square of side length $s$, the diagonal is $\\sqrt{2}s$, meaning the half-diagonal $CP$ is:\n $$CP = \\frac{s\\sqrt{2}}{2} = \\frac{s}{\\sqrt{2}}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{In } \\triangle OCP: \\quad \\htmlData{anchor=side1}{\\color{@sky}{CP^2}} + \\htmlData{anchor=side2}{\\color{@amber}{OC^2}} = \\htmlData{anchor=hyp}{\\color{@purple}{OP^2}}$"},{"latex":"$$\\htmlData{anchor=sub1}{\\color{@sky}{\\left(\\frac{s}{\\sqrt{2}}\\right)^2}} + \\htmlData{anchor=sub2}{\\color{@amber}{(h - K)^2}} = \\htmlData{anchor=sub3}{\\color{@purple}{K^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"side1"},{"color":"amber","style":"text-color","anchor":"side2"},{"color":"purple","style":"text-color","anchor":"hyp"}],"connections":[{"to":"sub1","from":"side1","color":"sky","label":null,"style":"solid"},{"to":"sub2","from":"side2","color":"amber","label":null,"style":"solid"},{"to":"sub3","from":"hyp","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Apply Pythagoras' theorem","description":"We apply the Pythagorean theorem to our right-angled triangle $\\triangle OCP$ to link $s$, $h$, and $K$."},{"type":"text","order":3,"title":"Rearrange to solve for s^2","content":"Now we expand the terms and solve for $s^2$:\n\n$$\\frac{s^2}{2} + (h^2 - 2Kh + K^2) = K^2$$\n\nSubtract $K^2$ from both sides of the equation:\n\n$$\\frac{s^2}{2} + h^2 - 2Kh = 0$$\n\nIsolate the $\\frac{s^2}{2}$ term:\n\n$$\\frac{s^2}{2} = 2Kh - h^2$$\n\nMultiply both sides by $2$ to express $s^2$ in terms of $K$ and $h$:\n\n$$s^2 = 4Kh - 2h^2 = 2(2Kh - h^2)$$\n\nThis completes the key algebraic step to represent the base area $s^2$."},{"type":"text","order":4,"title":"Substitute back into volume","content":"Now substitute our expression for $s^2$ back into the pyramid volume formula $V = \\frac{1}{3}s^2h$:\n\n$$\\begin{aligned} V &= \\frac{1}{3} \\left( 4Kh - 2h^2 \\right) h \\\\ &= \\frac{1}{3} \\left( 4Kh^2 - 2h^3 \\right) \\\\ &= \\frac{2}{3} (2Kh^2 - h^3) \\end{aligned}$$\n\nThis is the exact required expression for the volume of the pyramid."}]},{"id":"coordinate_geometry","label":"Coordinate Geometry","steps":[{"type":"text","order":0,"title":"Set up coordinate system","content":"Let's place the center of the solid crystal sphere at the origin $(0, 0, 0)$ of a 3D Cartesian coordinate system.\n\nThe equation of a sphere of radius $K$ centered at the origin is given by:\n\n$$x^2 + y^2 + z^2 = K^2$$\n\nSince the vertex of the pyramid and the center of its base lie on the same diameter, we can align this diameter with the $z$-axis. Let's place the top vertex of the pyramid at the topmost point of the sphere, which is at coordinates $(0, 0, K)$."},{"type":"text","order":1,"title":"Represent the base plane","content":"The pyramid has a vertical height $h$. Starting from the top vertex at $z = K$ and going vertically downwards, the center of the square base lies at a height of $h$ units below the vertex.\n\nTherefore, the $z$-coordinate of the base of the pyramid is:\n\n$$z = K - h$$"},{"type":"text","order":2,"title":"Find the circular cross-section","content":"The four vertices of the square base of the pyramid lie on the surface of the sphere. \n\nWe can find the radius $r$ of the horizontal circular cross-section of the sphere at $z = K - h$ by substituting $z$ into the equation of the sphere:\n\n$$x^2 + y^2 + (K - h)^2 = K^2$$\n\nThis can be written as:\n\n$$x^2 + y^2 = K^2 - (K - h)^2$$\n\nHere, $r^2 = x^2 + y^2$ is the square of the radius of the circular path on which the four base vertices lie:\n\n$$r^2 = K^2 - (K^2 - 2Kh + h^2) = 2Kh - h^2$$"},{"type":"text","order":3,"title":"Relate circle to base","content":"The base of the pyramid is a square of side length $s$. This square is inscribed within the circular cross-section of radius $r$.\n\nThe diagonal of this square is equal to the diameter of the circular cross-section, which is $2r$.\n\nUsing Pythagoras' Theorem on the square base, the relationship between the side length $s$ and the diagonal $2r$ is:\n\n$$s^2 + s^2 = (2r)^2$$\n\n$$2s^2 = 4r^2 \\implies s^2 = 2r^2$$"},{"type":"text","order":4,"title":"Express base area s^2","content":"Now we substitute our expression for $r^2$ from Step 2 into our relation for $s^2$:\n\n$$s^2 = 2(2Kh - h^2)$$\n\n$$s^2 = 4Kh - 2h^2$$"},{"type":"text","order":5,"title":"Calculate the pyramid volume","content":"The volume $V$ of a right square-based pyramid is given by the standard formula:\n\n$$V = \\frac{1}{3} \\times \\text{base area} \\times \\text{height} = \\frac{1}{3}s^2h$$\n\nSubstituting our expression for $s^2$:\n\n$$\\begin{aligned} V &= \\frac{1}{3}(4Kh - 2h^2)h \\\\ V &= \\frac{1}{3}(4Kh^2 - 2h^3) \\\\ V &= \\frac{2}{3}(2Kh^2 - h^3) \\end{aligned}$$\n\nThis completes the proof!"}]}],"generatedAt":"2026-06-09T07:59:02.054Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"07776508-cced-4897-9932-446741d079ee","label":"b","content":"Find the maximum possible volume of the pyramid in terms of \$K\$.\nFully justify your answer.","markscheme":"\n- Differentiates the volume with respect to height to obtain $\\frac{dV}{dh} = \\frac{8}{3}Kh - 2h^2$ M1\n- Sets $\\frac{dV}{dh} = 0$ and factorises $2h(\\frac{4}{3}K - h) = 0$ to find the stationary point $h = \\frac{4}{3}K$ (as $h > 0$) A1\n- Substitutes $h = \\frac{4}{3}K$ into the volume expression: $V = \\frac{2}{3}(2K(\\frac{4}{3}K)^2 - (\\frac{4}{3}K)^3)$ M1\n- Evaluates the volume to find $V_{max} = \\frac{64}{81}K^3$ A1\n- Finds $\\frac{d^2V}{dh^2} = \\frac{8}{3}K - 4h$ and evaluates at $h = \\frac{4}{3}K$ to get $-\\frac{8}{3}K$, concluding it is a maximum as $\\frac{d^2V}{dh^2} < 0$ M1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"calculus_height","label":"Calculus with respect to Height","steps":[{"type":"text","order":0,"title":"Recall the volume formula","content":"The volume of a right square-based pyramid with side length $s$ and height $h$ is given by:\n\n$$V = \\frac{1}{3}s^2h$$\n\nFrom the geometry of the pyramid inscribed inside a sphere of radius $K$, the relationship between the side length $s$ and height $h$ is:\n\n$$s^2 = 2h(2K - h)$$\n\nSubstituting this into our volume formula gives the volume as a function of height $h$:\n\n$$V = \\frac{2}{3}\\left(2Kh^2 - h^3\\right)$$\n\nwhere $0 < h < 2K$."},{"type":"text","order":1,"title":"Differentiate with respect to $h$","content":"To find the stationary points, we first differentiate the volume function $V(h)$ with respect to $h$:\n\n$$\\frac{dV}{dh} = \\frac{2}{3}\\left(4Kh - 3h^2\\right) = \\frac{8}{3}Kh - 2h^2$$\n\nThis derivative is key to locating where the volume is at a maximum."},{"type":"text","order":2,"title":"Find the critical point","content":"We set the derivative equal to zero to find the critical points:\n\n$$\\frac{8}{3}Kh - 2h^2 = 0$$\n\nFactorising out $2h$:\n\n$$2h\\left(\\frac{4}{3}K - h\\right) = 0$$\n\nSince $h > 0$ (as $0 < h < 2K$), the only valid solution is:\n\n$$h = \\frac{4}{3}K$$"},{"type":"text","order":3,"title":"Perform second derivative test","content":"To justify that $h = \\frac{4}{3}K$ yields a local maximum, we find the second derivative of $V$ with respect to $h$:\n\n$$\\frac{d^2V}{dh^2} = \\frac{8}{3}K - 4h$$\n\nSubstituting $h = \\frac{4}{3}K$ into the second derivative:\n\n$$\\left.\\frac{d^2V}{dh^2}\\right|_{h=\\frac{4}{3}K} = \\frac{8}{3}K - 4\\left(\\frac{4}{3}K\\right) = \\frac{8}{3}K - \\frac{16}{3}K = -\\frac{8}{3}K$$\n\nSince $K > 0$, we have $-\\frac{8}{3}K < 0$. \n\nBecause the second derivative is negative, $h = \\frac{4}{3}K$ indeed corresponds to a maximum volume."},{"type":"text","order":4,"title":"Calculate the maximum volume","content":"Finally, we substitute $h = \\frac{4}{3}K$ back into our volume equation to find the maximum volume:\n\n$$\\begin{aligned} V_{\\text{max}} &= \\frac{2}{3}\\left(2K\\left(\\frac{4}{3}K\\right)^2 - \\left(\\frac{4}{3}K\\right)^3\\right) \\\\ &= \\frac{2}{3}\\left(2K\\left(\\frac{16}{9}K^2\\right) - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{32}{9}K^3 - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{96}{27}K^3 - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{32}{27}K^3\\right) \\\\ &= \\frac{64}{81}K^3 \\end{aligned}$$\n\nThus, the maximum possible volume of the pyramid in terms of $K$ is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"}]},{"id":"calculus_distance","label":"Calculus with respect to Base Position","steps":[{"type":"text","order":0,"title":"Define the variable z","content":"Let's define $z$ as the signed distance from the center of the sphere of radius $K$ to the center of the square base of the pyramid, where $-K < z < K$.\n\nThe vertex of the pyramid is located on the sphere's surface at a distance of $K$ from the center along the same diameter. Therefore, the total vertical height $h$ of the pyramid is:\n\n$$h = K + z$$"},{"type":"text","order":1,"title":"Express base side length","content":"The square base is inscribed in a circular cross-section of the sphere at distance $|z|$ from the center.\n\nUsing Pythagoras' theorem, the radius $r$ of this circular cross-section is given by:\n\n$$r^2 = K^2 - z^2$$\n\nSince the square base of side length $s$ is inscribed in this circle, its diagonal is $2r$. Applying Pythagoras' theorem to the square base:\n\n$$s^2 + s^2 = (2r)^2 \\implies 2s^2 = 4r^2 \\implies s^2 = 2r^2$$\n\nSubstituting $r^2 = K^2 - z^2$ gives:\n\n$$s^2 = 2(K^2 - z^2)$$"},{"type":"text","order":2,"title":"Formulate volume as function","content":"The volume $V$ of a right square-based pyramid is given by the formula:\n\n$$V = \\frac{1}{3}s^2h$$\n\nSubstituting our expressions for $s^2$ and $h$ in terms of $z$:\n\n$$V(z) = \\frac{1}{3} \\cdot 2(K^2 - z^2)(K + z) = \\frac{2}{3}(K^2 - z^2)(K + z)$$\n\nExpanding this expression to make differentiation straightforward:\n\n$$\\begin{aligned} V(z) &= \\frac{2}{3}(K^3 + K^2z - Kz^2 - z^3) \\end{aligned}$$"},{"type":"text","order":3,"title":"Differentiate to find stationary point","content":"To find the maximum volume, we differentiate $V(z)$ with respect to $z$:\n\n$$\\frac{dV}{dz} = \\frac{2}{3}(K^2 - 2Kz - 3z^2)$$\n\nSetting $\\frac{dV}{dz} = 0$ to find the stationary points:\n\n$$3z^2 + 2Kz - K^2 = 0$$\n\nWe can factorise this quadratic equation:\n\n$$(3z - K)(z + K) = 0$$\n\nThis gives two solutions:\n\n$$z = \\frac{K}{3} \\quad \\text{or} \\quad z = -K$$\n\nSince $h = K + z$ must be strictly greater than $0$ ($h > 0$), we reject $z = -K$. Thus, the only valid stationary point is:\n\n$$z = \\frac{K}{3}$$\n\n(This corresponds to $h = K + \\frac{K}{3} = \\frac{4}{3}K$, which matches the standard height-based approach)."},{"type":"text","order":4,"title":"Justify the maximum using second derivative","content":"To confirm that $z = \\frac{K}{3}$ yields a maximum volume, we compute the second derivative of $V$ with respect to $z$:\n\n$$\\frac{d^2V}{dz^2} = \\frac{2}{3}(-2K - 6z)$$\n\nEvaluating this at $z = \\frac{K}{3}$:\n\n$$\\frac{d^2V}{dz^2}\\bigg|_{z = \\frac{K}{3}} = \\frac{2}{3}\\left(-2K - 6\\left(\\frac{K}{3}\\right)\\right) = \\frac{2}{3}(-4K) = -\\frac{8}{3}K$$\n\nSince $K > 0$, the second derivative $-\\frac{8}{3}K < 0$, which confirms that a maximum occurs at $z = \\frac{K}{3}$."},{"type":"text","order":5,"title":"Calculate maximum volume","content":"Now, we substitute $z = \\frac{K}{3}$ back into our volume formula:\n\n$$\\begin{aligned} V_{\\text{max}} &= \\frac{2}{3}\\left(K^2 - \\left(\\frac{K}{3}\\right)^2\\right)\\left(K + \\frac{K}{3}\\right) \\\\ &= \\frac{2}{3}\\left(K^2 - \\frac{K^2}{9}\\right)\\left(\\frac{4}{3}K\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{8}{9}K^2\\right)\\left(\\frac{4}{3}K\\right) \\\\ &= \\frac{64}{81}K^3 \\end{aligned}$$\n\nThus, the maximum possible volume of the pyramid in terms of $K$ is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"}]},{"id":"am_gm_inequality","label":"Using AM-GM Inequality","steps":[{"type":"text","order":0,"title":"Establish the volume formula","content":"A right pyramid with a square base of side length $s$ and height $h$ has a volume of:\n\n$$V = \\frac{1}{3}s^2h$$\n\nFrom the geometry of the sphere of radius $K$, the relationship between the base side length $s$ and the height $h$ is given by:\n\n$$s^2 = 2h(2K - h)$$\n\nSubstituting this into the volume formula gives:\n\n$$V = \\frac{2}{3}h^2(2K - h)$$\n\nwhere $0 < h < 2K$."},{"type":"text","order":1,"title":"Prepare terms for AM-GM","content":"To apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find the maximum value of $V$ without calculus, we rewrite the variable parts of the expression so that their sum is a constant. \n\nWe express the $h^2(2K - h)$ part using three factors: $\\frac{h}{2}$, $\\frac{h}{2}$, and $(2K - h)$.\n\nNotice that their sum is a constant:\n\n$$\\frac{h}{2} + \\frac{h}{2} + (2K - h) = 2K$$\n\nWe can rewrite the volume equation as:\n\n$$V = \\frac{8}{3}\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\\right)(2K - h)$$\n\nSince $\\frac{8}{3} \\times \\frac{1}{4} = \\frac{2}{3}$, this expression is equivalent to our original formula."},{"type":"text","order":2,"title":"Apply AM-GM inequality","content":"For any three positive real numbers $a$, $b$, and $c$, the AM-GM inequality states:\n\n$$\\frac{a + b + c}{3} \\geq \\sqrt[3]{abc} \\implies abc \\leq \\left(\\frac{a + b + c}{3}\\right)^3$$\n\nLet $a = \\frac{h}{2}$, $b = \\frac{h}{2}$, and $c = 2K - h$. Substituting these into the inequality gives:\n\n$$\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\right)(2K - h) \\leq \\left(\\frac{\\frac{h}{2} + \\frac{h}{2} + 2K - h}{3}\\right)^3$$\n\nSimplifying the right-hand side:\n\n$$\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\right)(2K - h) \\leq \\left(\\frac{2K}{3}\\right)^3 = \\frac{8}{27}K^3$$"},{"type":"text","order":3,"title":"Calculate the maximum volume","content":"Now we substitute this maximum value back into our rewritten expression for the volume $V$:\n\n$$V = \\frac{8}{3}\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\right)(2K - h) \\leq \\frac{8}{3} \\left(\\frac{8}{27}K^3\\right)$$\n\nEvaluating the multiplication:\n\n$$V \\leq \\frac{64}{81}K^3$$\n\nThus, the maximum possible volume of the pyramid is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"},{"type":"text","order":4,"title":"Justify the maximum condition","content":"Equality in the AM-GM inequality occurs if and only if all terms are equal:\n\n$$\\frac{h}{2} = 2K - h$$\n\nSolving for $h$:\n\n$$\\begin{aligned} h &= 4K - 2h \\\\ 3h &= 4K \\\\ h &= \\frac{4}{3}K \\end{aligned}$$\n\nSince $0 < \\frac{4}{3}K < 2K$, this height is valid and physically achievable within the sphere. This completely justifies that the maximum volume is indeed $V_{\\text{max}} = \\frac{64}{81}K^3$."}]}],"generatedAt":"2026-06-11T03:57:13.676Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"3f2a2356-b41c-4950-8022-379e25ef2de4:pythagorean","label":"(a) Pythagorean Theorem","steps":[{"type":"text","order":0,"title":"Recall the volume formula","content":"To find the volume $V$ of a right square-based pyramid, we use the standard formula:\n\n$$V = \\frac{1}{3} \\times \\text{base area} \\times \\text{height}$$\n\nSince the base of the pyramid is a square with side length $s$, its area is $s^2$. The vertical height is given as $h$. Substituting these into the formula yields:\n\n$$V = \\frac{1}{3}s^2h$$\n\nTo show that $V = \\frac{2}{3}(2Kh^2 - h^3)$, we need to express $s^2$ in terms of the sphere's radius $K$ and the pyramid's height $h$.","highlights":[{"text":"Show that the volume, \$V\$, of the pyramid is given by\n\\[V = \\frac{2}{3}(2Kh^2 - h^3)\\]","location":"specification"}]},{"type":"text","order":1,"title":"Define the right-angled triangle","content":"Let the center of the solid crystal sphere be $O$, the center of the pyramid's square base be $C$, and one of the vertices of the square base be $P$.\n\nWe can form a right-angled triangle $\\triangle OCP$:\n- The hypotenuse $OP$ is the radius of the sphere, so $OP = K$.\n- The side $OC$ is the vertical distance from the center of the sphere to the base of the pyramid. Since the total height of the pyramid is $h$ from the apex (which lies on the sphere's surface) to the base, and the radius from the apex to the sphere's center is $K$, this distance is given by:\n $$OC = |h - K|$$\n- The side $CP$ is the distance from the center of the square base to a corner vertex. For a square of side length $s$, the diagonal is $\\sqrt{2}s$, meaning the half-diagonal $CP$ is:\n $$CP = \\frac{s\\sqrt{2}}{2} = \\frac{s}{\\sqrt{2}}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{In } \\triangle OCP: \\quad \\htmlData{anchor=side1}{\\color{@sky}{CP^2}} + \\htmlData{anchor=side2}{\\color{@amber}{OC^2}} = \\htmlData{anchor=hyp}{\\color{@purple}{OP^2}}$"},{"latex":"$$\\htmlData{anchor=sub1}{\\color{@sky}{\\left(\\frac{s}{\\sqrt{2}}\\right)^2}} + \\htmlData{anchor=sub2}{\\color{@amber}{(h - K)^2}} = \\htmlData{anchor=sub3}{\\color{@purple}{K^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"side1"},{"color":"amber","style":"text-color","anchor":"side2"},{"color":"purple","style":"text-color","anchor":"hyp"}],"connections":[{"to":"sub1","from":"side1","color":"sky","label":null,"style":"solid"},{"to":"sub2","from":"side2","color":"amber","label":null,"style":"solid"},{"to":"sub3","from":"hyp","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Apply Pythagoras' theorem","description":"We apply the Pythagorean theorem to our right-angled triangle $\\triangle OCP$ to link $s$, $h$, and $K$."},{"type":"text","order":3,"title":"Rearrange to solve for s^2","content":"Now we expand the terms and solve for $s^2$:\n\n$$\\frac{s^2}{2} + (h^2 - 2Kh + K^2) = K^2$$\n\nSubtract $K^2$ from both sides of the equation:\n\n$$\\frac{s^2}{2} + h^2 - 2Kh = 0$$\n\nIsolate the $\\frac{s^2}{2}$ term:\n\n$$\\frac{s^2}{2} = 2Kh - h^2$$\n\nMultiply both sides by $2$ to express $s^2$ in terms of $K$ and $h$:\n\n$$s^2 = 4Kh - 2h^2 = 2(2Kh - h^2)$$\n\nThis completes the key algebraic step to represent the base area $s^2$."},{"type":"text","order":4,"title":"Substitute back into volume","content":"Now substitute our expression for $s^2$ back into the pyramid volume formula $V = \\frac{1}{3}s^2h$:\n\n$$\\begin{aligned} V &= \\frac{1}{3} \\left( 4Kh - 2h^2 \\right) h \\\\ &= \\frac{1}{3} \\left( 4Kh^2 - 2h^3 \\right) \\\\ &= \\frac{2}{3} (2Kh^2 - h^3) \\end{aligned}$$\n\nThis is the exact required expression for the volume of the pyramid."}]},{"id":"3f2a2356-b41c-4950-8022-379e25ef2de4:coordinate_geometry","label":"(a) Coordinate Geometry","steps":[{"type":"text","order":0,"title":"Set up coordinate system","content":"Let's place the center of the solid crystal sphere at the origin $(0, 0, 0)$ of a 3D Cartesian coordinate system.\n\nThe equation of a sphere of radius $K$ centered at the origin is given by:\n\n$$x^2 + y^2 + z^2 = K^2$$\n\nSince the vertex of the pyramid and the center of its base lie on the same diameter, we can align this diameter with the $z$-axis. Let's place the top vertex of the pyramid at the topmost point of the sphere, which is at coordinates $(0, 0, K)$."},{"type":"text","order":1,"title":"Represent the base plane","content":"The pyramid has a vertical height $h$. Starting from the top vertex at $z = K$ and going vertically downwards, the center of the square base lies at a height of $h$ units below the vertex.\n\nTherefore, the $z$-coordinate of the base of the pyramid is:\n\n$$z = K - h$$"},{"type":"text","order":2,"title":"Find the circular cross-section","content":"The four vertices of the square base of the pyramid lie on the surface of the sphere. \n\nWe can find the radius $r$ of the horizontal circular cross-section of the sphere at $z = K - h$ by substituting $z$ into the equation of the sphere:\n\n$$x^2 + y^2 + (K - h)^2 = K^2$$\n\nThis can be written as:\n\n$$x^2 + y^2 = K^2 - (K - h)^2$$\n\nHere, $r^2 = x^2 + y^2$ is the square of the radius of the circular path on which the four base vertices lie:\n\n$$r^2 = K^2 - (K^2 - 2Kh + h^2) = 2Kh - h^2$$"},{"type":"text","order":3,"title":"Relate circle to base","content":"The base of the pyramid is a square of side length $s$. This square is inscribed within the circular cross-section of radius $r$.\n\nThe diagonal of this square is equal to the diameter of the circular cross-section, which is $2r$.\n\nUsing Pythagoras' Theorem on the square base, the relationship between the side length $s$ and the diagonal $2r$ is:\n\n$$s^2 + s^2 = (2r)^2$$\n\n$$2s^2 = 4r^2 \\implies s^2 = 2r^2$$"},{"type":"text","order":4,"title":"Express base area s^2","content":"Now we substitute our expression for $r^2$ from Step 2 into our relation for $s^2$:\n\n$$s^2 = 2(2Kh - h^2)$$\n\n$$s^2 = 4Kh - 2h^2$$"},{"type":"text","order":5,"title":"Calculate the pyramid volume","content":"The volume $V$ of a right square-based pyramid is given by the standard formula:\n\n$$V = \\frac{1}{3} \\times \\text{base area} \\times \\text{height} = \\frac{1}{3}s^2h$$\n\nSubstituting our expression for $s^2$:\n\n$$\\begin{aligned} V &= \\frac{1}{3}(4Kh - 2h^2)h \\\\ V &= \\frac{1}{3}(4Kh^2 - 2h^3) \\\\ V &= \\frac{2}{3}(2Kh^2 - h^3) \\end{aligned}$$\n\nThis completes the proof!"}]},{"id":"07776508-cced-4897-9932-446741d079ee:calculus_height","label":"(b) Calculus with respect to Height","steps":[{"type":"text","order":0,"title":"Recall the volume formula","content":"The volume of a right square-based pyramid with side length $s$ and height $h$ is given by:\n\n$$V = \\frac{1}{3}s^2h$$\n\nFrom the geometry of the pyramid inscribed inside a sphere of radius $K$, the relationship between the side length $s$ and height $h$ is:\n\n$$s^2 = 2h(2K - h)$$\n\nSubstituting this into our volume formula gives the volume as a function of height $h$:\n\n$$V = \\frac{2}{3}\\left(2Kh^2 - h^3\\right)$$\n\nwhere $0 < h < 2K$."},{"type":"text","order":1,"title":"Differentiate with respect to $h$","content":"To find the stationary points, we first differentiate the volume function $V(h)$ with respect to $h$:\n\n$$\\frac{dV}{dh} = \\frac{2}{3}\\left(4Kh - 3h^2\\right) = \\frac{8}{3}Kh - 2h^2$$\n\nThis derivative is key to locating where the volume is at a maximum."},{"type":"text","order":2,"title":"Find the critical point","content":"We set the derivative equal to zero to find the critical points:\n\n$$\\frac{8}{3}Kh - 2h^2 = 0$$\n\nFactorising out $2h$:\n\n$$2h\\left(\\frac{4}{3}K - h\\right) = 0$$\n\nSince $h > 0$ (as $0 < h < 2K$), the only valid solution is:\n\n$$h = \\frac{4}{3}K$$"},{"type":"text","order":3,"title":"Perform second derivative test","content":"To justify that $h = \\frac{4}{3}K$ yields a local maximum, we find the second derivative of $V$ with respect to $h$:\n\n$$\\frac{d^2V}{dh^2} = \\frac{8}{3}K - 4h$$\n\nSubstituting $h = \\frac{4}{3}K$ into the second derivative:\n\n$$\\left.\\frac{d^2V}{dh^2}\\right|_{h=\\frac{4}{3}K} = \\frac{8}{3}K - 4\\left(\\frac{4}{3}K\\right) = \\frac{8}{3}K - \\frac{16}{3}K = -\\frac{8}{3}K$$\n\nSince $K > 0$, we have $-\\frac{8}{3}K < 0$. \n\nBecause the second derivative is negative, $h = \\frac{4}{3}K$ indeed corresponds to a maximum volume."},{"type":"text","order":4,"title":"Calculate the maximum volume","content":"Finally, we substitute $h = \\frac{4}{3}K$ back into our volume equation to find the maximum volume:\n\n$$\\begin{aligned} V_{\\text{max}} &= \\frac{2}{3}\\left(2K\\left(\\frac{4}{3}K\\right)^2 - \\left(\\frac{4}{3}K\\right)^3\\right) \\\\ &= \\frac{2}{3}\\left(2K\\left(\\frac{16}{9}K^2\\right) - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{32}{9}K^3 - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{96}{27}K^3 - \\frac{64}{27}K^3\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{32}{27}K^3\\right) \\\\ &= \\frac{64}{81}K^3 \\end{aligned}$$\n\nThus, the maximum possible volume of the pyramid in terms of $K$ is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"}]},{"id":"07776508-cced-4897-9932-446741d079ee:calculus_distance","label":"(b) Calculus with respect to Base Position","steps":[{"type":"text","order":0,"title":"Define the variable z","content":"Let's define $z$ as the signed distance from the center of the sphere of radius $K$ to the center of the square base of the pyramid, where $-K < z < K$.\n\nThe vertex of the pyramid is located on the sphere's surface at a distance of $K$ from the center along the same diameter. Therefore, the total vertical height $h$ of the pyramid is:\n\n$$h = K + z$$"},{"type":"text","order":1,"title":"Express base side length","content":"The square base is inscribed in a circular cross-section of the sphere at distance $|z|$ from the center.\n\nUsing Pythagoras' theorem, the radius $r$ of this circular cross-section is given by:\n\n$$r^2 = K^2 - z^2$$\n\nSince the square base of side length $s$ is inscribed in this circle, its diagonal is $2r$. Applying Pythagoras' theorem to the square base:\n\n$$s^2 + s^2 = (2r)^2 \\implies 2s^2 = 4r^2 \\implies s^2 = 2r^2$$\n\nSubstituting $r^2 = K^2 - z^2$ gives:\n\n$$s^2 = 2(K^2 - z^2)$$"},{"type":"text","order":2,"title":"Formulate volume as function","content":"The volume $V$ of a right square-based pyramid is given by the formula:\n\n$$V = \\frac{1}{3}s^2h$$\n\nSubstituting our expressions for $s^2$ and $h$ in terms of $z$:\n\n$$V(z) = \\frac{1}{3} \\cdot 2(K^2 - z^2)(K + z) = \\frac{2}{3}(K^2 - z^2)(K + z)$$\n\nExpanding this expression to make differentiation straightforward:\n\n$$\\begin{aligned} V(z) &= \\frac{2}{3}(K^3 + K^2z - Kz^2 - z^3) \\end{aligned}$$"},{"type":"text","order":3,"title":"Differentiate to find stationary point","content":"To find the maximum volume, we differentiate $V(z)$ with respect to $z$:\n\n$$\\frac{dV}{dz} = \\frac{2}{3}(K^2 - 2Kz - 3z^2)$$\n\nSetting $\\frac{dV}{dz} = 0$ to find the stationary points:\n\n$$3z^2 + 2Kz - K^2 = 0$$\n\nWe can factorise this quadratic equation:\n\n$$(3z - K)(z + K) = 0$$\n\nThis gives two solutions:\n\n$$z = \\frac{K}{3} \\quad \\text{or} \\quad z = -K$$\n\nSince $h = K + z$ must be strictly greater than $0$ ($h > 0$), we reject $z = -K$. Thus, the only valid stationary point is:\n\n$$z = \\frac{K}{3}$$\n\n(This corresponds to $h = K + \\frac{K}{3} = \\frac{4}{3}K$, which matches the standard height-based approach)."},{"type":"text","order":4,"title":"Justify the maximum using second derivative","content":"To confirm that $z = \\frac{K}{3}$ yields a maximum volume, we compute the second derivative of $V$ with respect to $z$:\n\n$$\\frac{d^2V}{dz^2} = \\frac{2}{3}(-2K - 6z)$$\n\nEvaluating this at $z = \\frac{K}{3}$:\n\n$$\\frac{d^2V}{dz^2}\\bigg|_{z = \\frac{K}{3}} = \\frac{2}{3}\\left(-2K - 6\\left(\\frac{K}{3}\\right)\\right) = \\frac{2}{3}(-4K) = -\\frac{8}{3}K$$\n\nSince $K > 0$, the second derivative $-\\frac{8}{3}K < 0$, which confirms that a maximum occurs at $z = \\frac{K}{3}$."},{"type":"text","order":5,"title":"Calculate maximum volume","content":"Now, we substitute $z = \\frac{K}{3}$ back into our volume formula:\n\n$$\\begin{aligned} V_{\\text{max}} &= \\frac{2}{3}\\left(K^2 - \\left(\\frac{K}{3}\\right)^2\\right)\\left(K + \\frac{K}{3}\\right) \\\\ &= \\frac{2}{3}\\left(K^2 - \\frac{K^2}{9}\\right)\\left(\\frac{4}{3}K\\right) \\\\ &= \\frac{2}{3}\\left(\\frac{8}{9}K^2\\right)\\left(\\frac{4}{3}K\\right) \\\\ &= \\frac{64}{81}K^3 \\end{aligned}$$\n\nThus, the maximum possible volume of the pyramid in terms of $K$ is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"}]},{"id":"07776508-cced-4897-9932-446741d079ee:am_gm_inequality","label":"(b) Using AM-GM Inequality","steps":[{"type":"text","order":0,"title":"Establish the volume formula","content":"A right pyramid with a square base of side length $s$ and height $h$ has a volume of:\n\n$$V = \\frac{1}{3}s^2h$$\n\nFrom the geometry of the sphere of radius $K$, the relationship between the base side length $s$ and the height $h$ is given by:\n\n$$s^2 = 2h(2K - h)$$\n\nSubstituting this into the volume formula gives:\n\n$$V = \\frac{2}{3}h^2(2K - h)$$\n\nwhere $0 < h < 2K$."},{"type":"text","order":1,"title":"Prepare terms for AM-GM","content":"To apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find the maximum value of $V$ without calculus, we rewrite the variable parts of the expression so that their sum is a constant. \n\nWe express the $h^2(2K - h)$ part using three factors: $\\frac{h}{2}$, $\\frac{h}{2}$, and $(2K - h)$.\n\nNotice that their sum is a constant:\n\n$$\\frac{h}{2} + \\frac{h}{2} + (2K - h) = 2K$$\n\nWe can rewrite the volume equation as:\n\n$$V = \\frac{8}{3}\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\\right)(2K - h)$$\n\nSince $\\frac{8}{3} \\times \\frac{1}{4} = \\frac{2}{3}$, this expression is equivalent to our original formula."},{"type":"text","order":2,"title":"Apply AM-GM inequality","content":"For any three positive real numbers $a$, $b$, and $c$, the AM-GM inequality states:\n\n$$\\frac{a + b + c}{3} \\geq \\sqrt[3]{abc} \\implies abc \\leq \\left(\\frac{a + b + c}{3}\\right)^3$$\n\nLet $a = \\frac{h}{2}$, $b = \\frac{h}{2}$, and $c = 2K - h$. Substituting these into the inequality gives:\n\n$$\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\\right)(2K - h) \\leq \\left(\\frac{\\frac{h}{2} + \\frac{h}{2} + 2K - h}{3}\\right)^3$$\n\nSimplifying the right-hand side:\n\n$$\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\\right)(2K - h) \\leq \\left(\\frac{2K}{3}\\right)^3 = \\frac{8}{27}K^3$$"},{"type":"text","order":3,"title":"Calculate the maximum volume","content":"Now we substitute this maximum value back into our rewritten expression for the volume $V$:\n\n$$V = \\frac{8}{3}\\left(\\frac{h}{2}\\right)\\left(\\frac{h}{2}\\right)(2K - h) \\leq \\frac{8}{3} \\left(\\frac{8}{27}K^3\\right)$$\n\nEvaluating the multiplication:\n\n$$V \\leq \\frac{64}{81}K^3$$\n\nThus, the maximum possible volume of the pyramid is:\n\n$$V_{\\text{max}} = \\frac{64}{81}K^3$$"},{"type":"text","order":4,"title":"Justify the maximum condition","content":"Equality in the AM-GM inequality occurs if and only if all terms are equal:\n\n$$\\frac{h}{2} = 2K - h$$\n\nSolving for $h$:\n\n$$\\begin{aligned} h &= 4K - 2h \\\\ 3h &= 4K \\\\ h &= \\frac{4}{3}K \\end{aligned}$$\n\nSince $0 < \\frac{4}{3}K < 2K$, this height is valid and physically achievable within the sphere. This completely justifies that the maximum volume is indeed $V_{\\text{max}} = \\frac{64}{81}K^3$."}]}],"generatedAt":"2026-06-11T03:57:13.808Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"a6a94817-fb09-44b2-b404-fe49164e1594","revisionId":"fd1a35b1-5d77-41cd-8869-ec37683ccfe0","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"hard","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"274dceb3-8404-4b8a-bdd5-58b77b6208de","label":null,"content":"For $x>0$, the curve $y=4x^2+16/x$ has a stationary point. 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Differentiation","topicId":"02df946e-8a41-4062-ae8d-11af60f06e7e","topicSidebar":"$L3c","topicSlug":"differentiation","topicTitle":"1.6 Differentiation"}]}],"$L3d","$L3e"]}]

1.6 Differentiation

Question 14

1.6 Differentiation Questions

1.6 Differentiation

Question 14

1.6 Differentiation Questions