Q.A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $2$ m and volume is $8$ m$^3$. If building of tank costs Rs $70$ per sq metres for the base and Rs $45$ per square metre for sides. What is the cost of least expensive tank?

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 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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6. 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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7. 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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8. 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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9. 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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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-2022-07-08-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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12. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.The least intercept made by a tangent to the ellipse $\dfrac{x^2}{64} + \dfrac{y^2}{49} = 1$ with coordinate axes is (A) $40$ (B) $10$ (C) $15$ (D) $100$

    ›Reveal solutionSolution

    The minimum length of the tangent segment cut off between the axes by a tangent to an ellipse equals $a+b$; here $a=8, b=7$ giving $15$.

    Concept and Intuition

    The tangent to the ellipse at the parametric point $(a\cos\theta, b\sin\theta)$ is $\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1$, which meets the x-axis at $x=a\sec\theta$ and the y-axis at $y=b\csc\theta$. As $\theta$ varies, this line segment's length changes — we must minimize it.

    Step-by-Step Solution

    1. Segment length squared: $L^2 = a^2\sec^2\theta + b^2\csc^2\theta$.
    2. Let $t=\sin^2\theta\in(0,1)$, so $\sec^2\theta = \dfrac{1}{1-t}$, $\csc^2\theta=\dfrac{1}{t}$: $L^2(t) = \dfrac{a^2}{1-t}+\dfrac{b^2}{t}$.
    3. Differentiate and set to zero: $\dfrac{a^2}{(1-t)^2} = \dfrac{b^2}{t^2} \Rightarrow at = b(1-t) \Rightarrow t=\dfrac{b}{a+b}$, so $1-t=\dfrac{a}{a+b}$. …

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