Q.A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is (A) $\frac{3r}{4}$ (B) $r^2$ (C) $\frac{r^2}{4}$ (D) $2r^2$

   ›Reveal solutionSolution

   The rectangle of maximum area that fits inside a circle is always a square; working through the calculus confirms the maximum area is $2r^2$.

   Concept and Intuition

   Every rectangle inscribed in a circle has its diagonal equal to the circle's diameter, since all four corners lie on the circle and the diagonal subtends the full circle. Among rectangles with a fixed diagonal, the one with maximum area is the square — this is the geometric content behind the calculus optimization below.

   Step-by-Step Solution

   1. Let the rectangle have half-width $x$ and half-height $y$ (centered at the circle's center), so its full diagonal equals the diameter $2r$: $x^2+y^2=r^2$.
   2. Area $A = (2x)(2y) = 4xy$.
   3. Maximize $A$ subject to $x^2+y^2=r^2$. Parametrize $x=r\cos\theta$, $y=r\sin\theta$, so $A=4r^2\sin\theta\cos\theta = 2r^2\sin2\theta$.
   4. $A$ is maximized when $\sin2\theta=1$, i.e. $\theta=\pi/4$, giving $A_{\max}=2r^2$. …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The maximum volume (in cu. m) of the right circular cone having slant height 3m. is (A) $6\pi$ (B) $3\sqrt{3}\,\pi$ (C) $\dfrac{4}{3}\pi$ (D) $2\sqrt{3}\,\pi$

   ›Reveal solutionSolution

   Express the cone's volume in terms of its height alone using the fixed slant height, then maximize; the maximum volume is $2\sqrt3\,\pi\ \text{m}^3$.

   Concept and Intuition

   With the slant height $l$ fixed, radius and height are linked by $r^2+h^2=l^2$ (Pythagoras on the cone's cross-section). This turns a two-variable optimization (over $r$ and $h$) into a single-variable one, which we handle with ordinary calculus.

   Step-by-Step Solution

   1. Given $l=3$, so $r^2+h^2=9\ \Rightarrow\ r^2=9-h^2$ (with $0<h<3$).
   2. Volume of a cone:

   $V=\frac13\pi r^2 h=\frac{\pi}{3}(9-h^2)h=\frac{\pi}{3}\left(9h-h^3\right).$

   3. Differentiate with respect to $h$ and set to zero:

   $\frac{dV}{dh}=\frac{\pi}{3}(9-3h^2)=0\ \Rightarrow\ h^2=3\ \Rightarrow\ h=\sqrt3.$

   4. Check it's a maximum: $\dfrac{d^2V}{dh^2}=\dfrac{\pi}{3}(-6h)<0$ for $h>0$, confirming a maximum.
   5. Substitute back: …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If a sector of maximum area is made with a wire of length $40\ cm$, then the area (in sq cms) of that sector is (A) $50$ (B) $100$ (C) $25$ (D) $200$

   ›Reveal solutionSolution

   Maximize sector area subject to a fixed wire (perimeter = two radii + arc) using single-variable calculus.

   Concept and Intuition

   A sector's boundary made of wire consists of two straight radii plus the curved arc, so the constraint is $2r+s=40$ where $s=r\theta$ is the arc length. The area formula $A=\tfrac12rs$ (half the product of radius and arc length) then becomes a function of $r$ alone once $s$ is eliminated via the constraint, letting ordinary optimization find the maximizing radius.

   Step-by-Step Solution

   1. Constraint: $2r+s=40\Rightarrow s=40-2r$ (with $0<r<20$).
   2. Area: $A=\dfrac12rs=\dfrac12r(40-2r)=20r-r^2$.
   3. Differentiate: $\dfrac{dA}{dr}=20-2r$. Set to zero: $r=10$.
   4. Check it's a maximum: $\dfrac{d^2A}{dr^2}=-2<0$, confirming a maximum. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the area of a right angled triangle with hypotenuse 5 is maximum, then its perimeter is (A) 12 (B) $2\sqrt{3}+\sqrt{13}+5$ (C) $7+\sqrt{21}$ (D) $5(\sqrt{2}+1)$

   ›Reveal solutionSolution

   This tests the AM–GM / Lagrange idea that for a fixed hypotenuse, the right triangle of maximum area is isosceles; the perimeter then works out to $5(\sqrt2+1)$.

   Concept and Intuition

   For a right triangle with hypotenuse $h$ fixed, $a^2+b^2=h^2$ is constant. The area is $\tfrac12 ab$, so maximizing area means maximizing the product $ab$ subject to a fixed sum of squares. By AM–GM, $\dfrac{a^2+b^2}{2}\ge ab$, with equality exactly when $a=b$. So the symmetric (isosceles) right triangle always has the largest area for a given hypotenuse — no calculus needed, though calculus confirms it.

   Step-by-Step Solution

   1. Let the legs be $a,b$ with $a^2+b^2=25$ (hypotenuse $=5$).
   2. Area $=\tfrac12 ab$ is maximized when $a=b$ (AM–GM equality case), i.e. $2a^2=25 \Rightarrow a=b=\dfrac{5}{\sqrt2}$. …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If P$(\alpha, \beta)$ is a point on the curve $9x^2 + 4y^2 = 144$ in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at P with the coordinate axis is S, then (A) $S = \sqrt{\alpha\beta}$ (B) $S = \alpha\beta$ (C) $S = 2\sqrt{\alpha\beta}$ (D) $S = 2\alpha\beta$

   ›Reveal solutionSolution

   The tangent-line intercept triangle's area, minimized over the ellipse, works out to exactly twice the product of the point's coordinates.

   Concept and Intuition

   The ellipse $9x^2+4y^2=144$ is $\dfrac{x^2}{16}+\dfrac{y^2}{36}=1$. The tangent at any point cuts the axes to form a right triangle whose area depends on where on the ellipse you are; calculus (or AM–GM) finds where that area is smallest, and then we check what algebraic relation the minimizing point's coordinates satisfy with that minimum area.

   Step-by-Step Solution

   1. Ellipse: $\dfrac{x^2}{16}+\dfrac{y^2}{36}=1$. Tangent at $P(\alpha,\beta)$: $\dfrac{x\alpha}{16}+\dfrac{y\beta}{36}=1$.
   2. Intercepts: $x$-intercept $=\dfrac{16}{\alpha}$, $y$-intercept $=\dfrac{36}{\beta}$.
   3. Triangle area: $S=\dfrac12\cdot\dfrac{16}{\alpha}\cdot\dfrac{36}{\beta}=\dfrac{288}{\alpha\beta}$.
   4. Parametrize $\alpha=4\cos\theta,\ \beta=6\sin\theta$ ($0<\theta<\pi/2$ for the first quadrant): $S=\dfrac{288}{24\sin\theta\cos\theta}=\dfrac{24}{\sin2\theta}$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If a running track of 500 ft. is to be laid out enclosing a playground, the shape of which is a rectangle with a semicircle at each end, then the length of the rectangular portion such that the area of the rectangular portion is to be maximum is (in feet). (A) $100$ (B) $125$ (C) $150$ (D) $200$

   ›Reveal solutionSolution

   This is a classic optimization problem: maximize the rectangular area of a stadium-shaped track of fixed perimeter. Answer: $x=125$ ft.

   Concept and Intuition

   The track's total perimeter is fixed at 500 ft. The two semicircular ends together form one full circle, and the two straight sides form the rectangle's length. Expressing the rectangle's area purely in terms of its length $x$ (using the perimeter constraint to eliminate the radius) turns this into a single-variable calculus optimization.

   Step-by-Step Solution

   1. Let the rectangle have length $x$ and width $2r$ (so the semicircles at each end have radius $r$).
   2. Total track length: two straight sides of length $x$ each, plus two semicircles (radius $r$) which together make one full circle of circumference $2\pi r$: $\;2x + 2\pi r = 500 \Rightarrow x+\pi r=250 \Rightarrow r=\dfrac{250-x}{\pi}$.
   3. Area of the rectangular portion: $A = x\cdot 2r = 2x\cdot\dfrac{250-x}{\pi} = \dfrac{500x-2x^2}{\pi}$. …

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7. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Given that the solid obtained by rotating a rectangle about one of its side is a cylinder. If the perimeter of a rectangle is 48 cm and the volume of the cylinder formed by rotating it is maximum, then the dimensions of that rectangle is (A) 14, 10 (B) 20, 4 (C) 18, 6 (D) 8, 16

   ›Reveal solutionSolution

   Maximise $V=\pi r^2 h$ subject to $2(r+h)=48$; the optimum rectangle is $8\times 16$.

   Concept and Intuition

   Rotating a rectangle about one of its sides sweeps the opposite side around in a circle, producing a cylinder whose height equals the rotation-axis side and whose radius equals the other side. This converts a plane geometry optimisation into a single-variable calculus problem once the perimeter constraint eliminates one variable.

   Step-by-Step Solution

   1. Let the side about which we rotate be $h$ (height of the cylinder) and the other side be $r$ (radius of the cylinder).
   2. Perimeter constraint: $2(h+r)=48 \Rightarrow h+r=24 \Rightarrow h=24-r$.
   3. Volume: $V(r)=\pi r^2 h = \pi r^2(24-r)=\pi(24r^2-r^3)$.
   4. $\dfrac{dV}{dr}=\pi(48r-3r^2)=3\pi r(16-r)$. Setting this to zero: $r=0$ (rejected, degenerate) or $r=16$.
   5. Then $h=24-16=8$. …

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8. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If the height of a cone of greatest volume that can be inscribed in a sphere of radius R is kR, then ratio of the volume of the cone to the volume of the sphere is (A) $8:27$ (B) $27:64$ (C) $8:125$ (D) $4:5$

   ›Reveal solutionSolution

   Standard optimization: the cone of greatest volume inscribed in a sphere has height $4R/3$, and its volume is $8/27$ of the sphere's.

   Concept and Intuition

   Setting the base circle of the cone at height $h$ above the sphere's lowest point (with apex at that lowest point), the base radius satisfies $r^2=2Rh-h^2$ by the geometry of the circle. Maximizing $V(h)=\frac13\pi r^2h$ gives the classical height $\frac{4R}{3}$.

   Step-by-Step Solution

   1. $r^2=R^2-(h-R)^2=2Rh-h^2$.
   2. $V(h)=\frac13\pi r^2h=\frac13\pi(2Rh^2-h^3)$.
   3. $\frac{dV}{dh}=\frac13\pi(4Rh-3h^2)=\frac13\pi h(4R-3h)$; setting this to $0$ (excluding $h=0$) gives $h=\frac{4R}{3}$, so $k=\frac43$.
   4. At this $h$: $r^2=2R\cdot\frac{4R}{3}-\left(\frac{4R}{3}\right)^2=\frac{8R^2}{3}-\frac{16R^2}{9}=\frac{8R^2}{9}$.
   5. $V_{max}=\frac13\pi\cdot\frac{8R^2}{9}\cdot\frac{4R}{3}=\frac{32}{81}\pi R^3$.
   6. $V_{sphere}=\frac43\pi R^3$. …

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9. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.A closed cylinder of given volume will have least surface area when the ratio of its height and base radius is (A) $2:1$ (B) $1:2$ (C) $2:3$ (D) $3:2$

   ›Reveal solutionSolution

   Minimising the total surface area of a closed cylinder for a fixed volume, via ordinary calculus, gives the classic result $h=2r$, i.e. height-to-radius ratio $2:1$.

   Concept and Intuition

   This is a standard optimisation problem: express the surface area as a function of one variable (using the volume constraint to eliminate the other), then find where its derivative vanishes.

   Step-by-Step Solution

   1. Volume constraint: $V=\pi r^2 h \Rightarrow h = \dfrac{V}{\pi r^2}$.
   2. Total surface area (closed cylinder, both circular ends included): $S = 2\pi r^2 + 2\pi r h$.
   3. Substitute $h$: $S = 2\pi r^2 + 2\pi r\cdot\dfrac{V}{\pi r^2} = 2\pi r^2 + \dfrac{2V}{r}$.
   4. Differentiate with respect to $r$: $\dfrac{dS}{dr} = 4\pi r - \dfrac{2V}{r^2}$.
   5. Set to zero: $4\pi r = \dfrac{2V}{r^2} \Rightarrow 4\pi r^3 = 2V \Rightarrow r^3 = \dfrac{V}{2\pi}$.
   6. Since $V=\pi r^2 h$: $r^3 = \dfrac{\pi r^2 h}{2\pi} = \dfrac{r^2h}{2} \Rightarrow r = \dfrac{h}{2} \Rightarrow h = 2r$. …

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10. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.The maximum area of a right angled triangle with hypotenuse h is (A) $\dfrac{h^2}{2\sqrt{2}}$ (B) $\dfrac{h^2}{2}$ (C) $\dfrac{h^2}{\sqrt{2}}$ (D) $\dfrac{h^2}{4}$

    ›Reveal solutionSolution

    Maximising the area of a right triangle with fixed hypotenuse $h$ occurs at the isosceles case, giving area $h^2/4$.

    Concept and Intuition

    For a right triangle with legs $a,b$ and fixed hypotenuse $h$ (so $a^2+b^2=h^2$ is a constraint), the area $\tfrac12 ab$ is maximised by symmetry when $a=b$ — this is a classic constrained-optimisation result, provable via calculus or the AM-GM inequality ($a^2+b^2\ge 2ab$, so $ab\le \tfrac{a^2+b^2}{2}=\tfrac{h^2}{2}$, with equality iff $a=b$).

    Step-by-Step Solution

    1. Let the legs be $a$ and $b=\sqrt{h^2-a^2}$ (from Pythagoras), and area $S=\tfrac12 a\sqrt{h^2-a^2}$.
    2. Maximise $S^2 = \tfrac14 a^2(h^2-a^2)$ instead (avoids the square root). Let $u=a^2$: $S^2=\tfrac14 u(h^2-u)$, a downward parabola in $u$, maximised at $u=h^2/2$.
    3. So $a^2=h^2/2 \Rightarrow a=h/\sqrt2$, and then $b^2=h^2-a^2=h^2/2 \Rightarrow b=h/\sqrt2$ too — the triangle is isosceles right-angled. …

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11. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.If an open cylinder of given surface area has maximum volume then its radius is (A) Height of the cylinder (B) Height of the cylinder / 2 (C) 2 times Height of the cylinder (D) 3 times Height of the cylinder

    ›Reveal solutionSolution

    A constrained-optimisation problem: maximise the volume of an open cylinder for a fixed surface area. Answer: the radius equals the height ($r=h$).

    Concept and Intuition

    "Open" cylinder means it has only one circular base (like a cup, no lid), so its total surface area is base + lateral surface, $S=\pi r^2+2\pi rh$ — different from a closed cylinder (two bases) which would give a different optimum ($h=2r$). Fixing $S$ lets you express $h$ in terms of $r$, turning the volume into a single-variable function of $r$ to maximise.

    Step-by-Step Solution

    1. Surface constraint: $S=\pi r^2+2\pi rh\Rightarrow h=\dfrac{S-\pi r^2}{2\pi r}$.
    2. Volume: $V=\pi r^2 h=\pi r^2\cdot\dfrac{S-\pi r^2}{2\pi r}=\dfrac{r(S-\pi r^2)}2=\dfrac{Sr-\pi r^3}2$.
    3. Differentiate w.r.t. $r$ and set to zero: $\dfrac{dV}{dr}=\dfrac{S-3\pi r^2}{2}=0\Rightarrow S=3\pi r^2$. …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.The perimeter of a sector is constant. If its area is to be maximum, the sectorical angle should be (A) $\dfrac{\pi^c}{6}$ (B) $\dfrac{\pi^c}{4}$ (C) $4^c$ (D) $2^c$

    ›Reveal solutionSolution

    Expressing the sector's area purely in terms of its radius (using the fixed-perimeter constraint) and maximizing gives $r=P/4$, and back-substituting gives the optimal sectorial angle $\theta=2$ radians.

    Concept and Intuition

    A circular sector's boundary consists of two straight radii and one arc, so its perimeter is $P=2r+r\theta$. Since $P$ is held fixed, $\theta$ is not a free variable — it is determined by $r$. This turns the two-variable area formula $A=\tfrac12r^2\theta$ into a single-variable optimisation problem in $r$ alone.

    Step-by-Step Solution

    1. Perimeter: $P=2r+r\theta$ (constant) $\Rightarrow \theta=\dfrac{P-2r}{r}$.
    2. Area: $A=\dfrac12r^2\theta=\dfrac12r^2\cdot\dfrac{P-2r}{r}=\dfrac12r(P-2r)=\dfrac{Pr}{2}-r^2$.
    3. Differentiate w.r.t. $r$: $\dfrac{dA}{dr}=\dfrac{P}{2}-2r$. Set to zero: $r=\dfrac{P}{4}$.
    4. Second derivative $\dfrac{d^2A}{dr^2}=-2<0$, confirming a maximum.
    5. Substitute back into $\theta=\dfrac{P-2r}{r}$: with $r=P/4$, $\theta=\dfrac{P-P/2}{P/4}=\dfrac{P/2}{P/4}=2$. …

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