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Q.A man has an expensive square-shaped piece of golden board of side 36 cm. He wants to turn it into a box without top by cutting a square from each corner and folding the flaps. Let xx cm be the side of square, which is cut from each corner. The figure shows a square sheet of side 3636 cm with a small square of side xx cm cut from each corner; folding up the four flaps forms an open box of height xx. Based on the above information, answer the following questions :

(i) Find the expression for the volume (V) of open box in terms of xx.
(ii) Find dVdx\dfrac{dV}{dx}.
(iii) Find the value of xx for which the volume (V) is maximum.
(OR)
(iii) Find the maximum volume of the open box.
CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★
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Open box: base side 36−2x36-2x, height xx, so V=x(36−2x)2V = x(36-2x)^2. dVdx=12(18−x)(6−x)=0\tfrac{dV}{dx} = 12(18-x)(6-x) = 0 gives x=6x=6 (rejecting x=18x=18); V′′<0V''<0 there, and Vmax⁡=6(24)2=3456 cm3V_{\max} = 6(24)^2 = 3456\ \text{cm}^3.

Volume of a cuboid (open box) =base area×height= \text{base area} \times \text{height}. A maximum occurs where dVdx=0\dfrac{dV}{dx} = 0 and d2Vdx2<0\dfrac{d^2V}{dx^2} < 0.

  1. Volume expression. Cutting squares of side xx from each corner of the 36×3636\times36 sheet leaves a base of side (36−2x)(36 - 2x); folding gives height xx.

    V=x(36−2x)2V = x(36 - 2x)^2

  2. Derivative.
  1. dVdx=(36−2x)2+x⋅2(36−2x)(−2)=(36−2x)2−4x(36−2x)\dfrac{dV}{dx} = (36-2x)^2 + x\cdot 2(36-2x)(-2) = (36-2x)^2 - 4x(36-2x).
  2. Factor (36−2x)(36-2x): =(36−2x)[(36−2x)−4x]=(36−2x)(36−6x)= (36-2x)\big[(36-2x) - 4x\big] = (36-2x)(36-6x).
  3. =12(18−x)(6−x)= 12(18 - x)(6 - x).

dVdx=12(18−x)(6−x)\dfrac{dV}{dx} = 12(18 - x)(6 - x)

(iii) Maximising xx.

  1. Set dVdx=0\dfrac{dV}{dx} = 0: 12(18−x)(6−x)=0⇒x=1812(18-x)(6-x) = 0 \Rightarrow x = 18 or x=6x = 6. …

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