Q.A metal box with a square base and vertical sides is to contain $1024$ cm$^3$. The material for the top and bottom costs Rs $5$/cm$^2$ and the material for the sides costs Rs $2.50$/cm$^2$. Find the least cost of the box.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.From a rectangular sheet having dimensions $30\ cm \times 80\ cm$, four equal squares of side $x\ cm$ are cut at each corner. The remaining sides of the rectangle are folded up vertically so as to form an open rectangular box. Find the value of 'x' for which the volume of the box formed is maximum. (A) $x = 30\ cm$ (B) $x = 20\ cm$ (C) $x = \dfrac{20}{3}\ cm$ (D) $x = 15\ cm$

   ›Reveal solutionSolution

   Express box volume as a function of the cut-square side $x$, maximize with calculus, and discard the root that isn't physically valid. The answer is (C).

   Concept and Intuition

   Cutting squares of side $x$ from each corner and folding up gives a box of dimensions $(30-2x)\times(80-2x)\times x$. The volume is a cubic in $x$ with two critical points; physically $x$ must be less than half the shorter side (15 cm), which rules out one root.

   Step-by-Step Solution

   1. $V(x) = x(30-2x)(80-2x)$.
   2. Expand: $(30-2x)(80-2x) = 2400 - 60x - 160x + 4x^2 = 2400 - 220x + 4x^2$.
   3. $V(x) = 2400x - 220x^2 + 4x^3$.
   4. $V'(x) = 2400 - 440x + 12x^2$. Set to $0$: $12x^2 - 440x + 2400 = 0 \Rightarrow 3x^2 - 110x + 600 = 0$.
   5. $x = \dfrac{110\pm\sqrt{110^2-4(3)(600)}}{6} = \dfrac{110\pm\sqrt{12100-7200}}{6} = \dfrac{110\pm70}{6}$, giving $x=30$ or $x=\dfrac{20}{3}$. …

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2. 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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3. 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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4. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.A window is in the shape of a rectangle, with a semi-circle fused to one of its sides, as shown in the figure. [FIGURE] (a rectangle with a semi-circle attached to its right side, forming a window shape) If the perimeter of the window is fixed as 20 units, then its maximum area can be _____ sq. units. (A) $\dfrac{400}{\pi + 4}$ (B) $\dfrac{20}{\pi + 4}$ (C) $\dfrac{100}{\pi + 4}$ (D) $\dfrac{200}{\pi + 4}$

   ›Reveal solutionSolution

   This is a constrained optimization problem: maximize the area of a rectangle with a semicircle on one side, given a fixed perimeter of 20. The maximum area is $\frac{200}{\pi+4}$, so the correct option is (D).

   We have a window shaped like a rectangle with a semicircle attached to its right side. The semicircle’s diameter equals the height of the rectangle. The total perimeter is fixed at 20 units. We want the maximum possible area.

   Why this approach works:

   When a shape’s perimeter is fixed, the area is maximized by making the shape as “round” as possible — but here the shape is partly rectangular, so we need to balance the rectangle’s width and height. We’ll express area in terms of one variable, then use calculus (or completing the square) to find the maximum.

   ---

   1. Define variables

      Let the rectangle have width $x$ (horizontal side) and height $y$ (vertical side). The semicircle sits on the right side, so its diameter is $y$, and its radius is $r = y/2$.

   2. Write the perimeter

      The perimeter consists of:

      • Left vertical side: $y$
      • Top horizontal side: $x$
      • Bottom horizontal side: $x$
      • Right vertical side: $y$ (but this is not part of the outer boundary — the semicircle replaces it)
      • The curved semicircular arc: $\pi r = \pi (y/2)$

      So total perimeter:

   $P = y + x + x + \pi \frac{y}{2} = 2x + y + \frac{\pi y}{2}$

   Given $P = 20$:

   $2x + y\left(1 + \frac{\pi}{2}\right) = 20$

   3. Solve for $x$ in terms of $y$

   $2x = 20 - y\left(1 + \frac{\pi}{2}\right) \quad\Rightarrow\quad x = 10 - \frac{y}{2}\left(1 + \frac{\pi}{2}\right)$

   Simplify:

   $x = 10 - \frac{y}{2} - \frac{\pi y}{4}$

   4. Write the area

      Area = rectangle area + semicircle area:

      • Rectangle: $x \cdot y$
      • Semicircle: $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{y}{2}\right)^2 = \frac{\pi y^2}{8}$

      So:

   $A = x y + \frac{\pi y^2}{8}$

   Substitute $x$:

   $A = y\left(10 - \frac{y}{2} - \frac{\pi y}{4}\right) + \frac{\pi y^2}{8}$

   $A = 10y - \frac{y^2}{2} - \frac{\pi y^2}{4} + \frac{\pi y^2}{8}$

   5. Combine the $y^2$ terms $-\frac{\pi y^2}{4} + \frac{\pi y^2}{8} = -\frac{\pi y^2}{8}$ So:

   $A = 10y - \frac{y^2}{2} - \frac{\pi y^2}{8}$

   Factor $y^2$:

   $A = 10y - y^2\left(\frac{1}{2} + \frac{\pi}{8}\right)$

   Write $\frac{1}{2} = \frac{4}{8}$, so:

   $A = 10y - y^2\left(\frac{4 + \pi}{8}\right)$

   6. Maximize using calculus …

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5. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.Maximum area of the rectangle inscribed in a circle of radius 10 cms is (A) $100$ (B) $200$ (C) $250$ (D) $150$

   ›Reveal solutionSolution

   This tests the classical optimization result that the square is the area-maximizing rectangle inscribed in a given circle.

   Concept and Intuition

   A rectangle inscribed in a circle of radius $r$ has its diagonal equal to the circle's diameter $2r$. If the sides are $x,y$, then $x^2+y^2=(2r)^2$, and by AM-GM, $xy$ (the area) is maximized when $x=y$, i.e. when the rectangle is a square.

   Step-by-Step Solution

   1. Let the rectangle have sides $x,y$ with diagonal $=2r=20$: $x^2+y^2=400$.
   2. Area $A=xy$. Maximize subject to $x^2+y^2=400$: by AM-GM/symmetry, max occurs at $x=y$.
   3. Then $2x^2=400\Rightarrow x^2=200 \Rightarrow x=y=\sqrt{200}$.
   4. Max area $=xy=200$. …

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6. 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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7. 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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8. 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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9. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Find the equation of a line passing through the point $(4,3)$, which cuts a triangle of minimum area from the first quadrant. (A) $3x+4y=24$ (B) $2x-y=5$ (C) $2x+y=8$ (D) $x-2y=5$

   ›Reveal solutionSolution

   A well-known optimization result: the line through a fixed interior point cutting the least-area triangle from the axes is the one for which that point bisects the intercepted segment.

   Concept and Intuition

   Let the line be $\dfrac{x}{p}+\dfrac{y}{q}=1$ passing through $(4,3)$: $\dfrac4p+\dfrac3q=1$. The triangle area is $\dfrac12 pq$. Minimizing area subject to this constraint (via AM-GM or calculus) shows the optimum occurs precisely when $(4,3)$ is the midpoint of the intercepts, i.e., $p=8, q=6$.

   Step-by-Step Solution

   1. General line through $(4,3)$ in intercept form: $\dfrac{x}{p}+\dfrac{y}{q}=1$ with $\dfrac{4}{p}+\dfrac{3}{q}=1$.
   2. The standard result for minimum-area triangle: the given point bisects the segment between the intercepts, so $p=2(4)=8$ and $q=2(3)=6$. …

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10. 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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11. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.Through the point $(4, 5)$, a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the x and y axes is ________ (A) $1 : 1$ (B) $3 : 4$ (C) $4 : 5$ (D) $2 : 3$

    ›Reveal solutionSolution

    Minimize the intercept-triangle area subject to the line passing through a fixed point, using single-variable calculus. Answer: intercept ratio $4:5$.

    Concept and Intuition

    The family of lines through $(4,5)$ with positive intercepts $a$ (on x-axis) and $b$ (on y-axis) satisfies $\frac4a+\frac5b=1$. As the line rotates through the fixed point, the triangle area $\frac12 ab$ changes; we minimize it using the constraint to reduce to one variable.

    Step-by-Step Solution

    1. Line: $\dfrac{x}{a}+\dfrac{y}{b}=1$ through $(4,5)$: $\dfrac4a+\dfrac5b=1$.
    2. Solve for $b$: $\dfrac5b = 1-\dfrac4a=\dfrac{a-4}{a}\Rightarrow b=\dfrac{5a}{a-4}$ (need $a>4$).
    3. Area function: $A(a)=\dfrac12 ab = \dfrac12\cdot a\cdot\dfrac{5a}{a-4}=\dfrac{5a^2}{2(a-4)}$. …

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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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