A box open at the top is made by cutting equal squares from the four corners of a 6× 6 m sheet and folding up the sides. The height of the box for maximum volume is:

D. None of these. Step 1: Cut squares of side x. The box has a square base of side 6-2x and height x: V=x(6-2x)^2, 0<x<3. Step 2: V′=(6-2x)^2-4x(6-2x)=(6-2x)(6-6x)=0 gives x=1 ( x=3 leaves no base). Step 3: V(1)=1× 4^2=16 cubic m is the maximum, so the height is 1 m, which is not in (A), (B) or (C). Check: V(0.9)=0.9× 4.2^2=15.876 and V(1.1)=1.1× 3.8^2=15.884, both below 16. Trap: (A) 2 m gives V(2)=2× 2^2=8 cubic m, only half the maximum.

Question 19 of 34 · Applications of Derivatives

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A box open at the top is made by cutting equal squares from the four corners of a $6 \times 6$ m sheet and folding up the sides. The height of the box for maximum volume is:

Answer (Detailed Solution Below)View answer ▾

Answer: Option D — None of these

Detailed Solution

Cut squares of side $x$ . The box has a square base of side $6 - 2 x$ and height $x$ : $V = x (6 - 2 x)^{2}$ , $0 < x < 3$ .

$V' = (6 - 2 x)^{2} - 4 x (6 - 2 x) = (6 - 2 x) (6 - 6 x) = 0$ gives $x = 1$ ( $x = 3$ leaves no base).

$V (1) = 1 \times 4^{2} = 16$ cubic m is the maximum, so the height is 1 m, which is not in (A), (B) or (C).

✓ Check

V (0.9) = 0.9 \times 4.2^{2} = 15.876

and

V (1.1) = 1.1 \times 3.8^{2} = 15.884

, both below 16.

⚠ Common trap

(A) 2 m gives

V (2) = 2 \times 2^{2} = 8

cubic m, only half the maximum.

Quick tip

$V = x (a - 2 x)^{2} is maximum at x = \frac{a}{6} = 1$

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Applications of Derivatives — Preparation

This question tests your understanding of Applications of Derivatives, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

This topic uses derivatives to solve practical and geometric problems. It covers the equations of tangents and normals to a curve, the rate of change of quantities (such as how fast the area of a circle grows as its radius increases), approximations using differentials, and increasing and decreasing functions. It also covers maxima and minima, using the first derivative test and the second derivative test, and word problems on optimisation such as finding the largest area for a fixed perimeter. Rolle's theorem and the mean value theorem are included as well. For maxima and minima, set f'(x) = 0 to find the critical points, then check f''(x): a negative value means a maximum and a positive value means a minimum. Remember that the slope of the tangent at a point is f'(x) and the slope of the normal is -1/f'(x). Practise the applications of derivatives questions below, each with a step-by-step solution.

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A box open at the top is made by cutting equal squares from the four corners of a $6 \times 6$ m sheet and folding up the sides. The height of the box for maximum volume is:

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A box open at the top is made by cutting equal squares from the four corners of a 6×6 m sheet and folding up the sides. The height of the box for maximum volume is:

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A box open at the top is made by cutting equal squares from the four corners of a $6 \times 6$ m sheet and folding up the sides. The height of the box for maximum volume is: