A closed pan with a square base and vertical sides can hold 64 cubic metres of water. For the total surface area to be minimum, the dimensions of the pan in metres are

4,4,4

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4,6,2

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4,4,2

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2,2,2

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Solution

The correct option is A 4,4,4
$V o l u m e = a^{2} h$
$64 = a^{2} h$
$T \cdot S \cdot A = 2 a^{2} + 4 a h$
$= 2 a^{2} + \frac{4 \times 64}{a}$
$\frac{d (T \cdot S \cdot A)}{d a} = 4 a - \frac{4 \times 64}{a^{2}}$
$a^{3} = 64$
$a = 4$ $h = 4$

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A closed pan with a square base and vertical sides can hold 64 cubic metres of water. For the total surface area to be minimum, the dimensions of the pan in metres are

The correct option is A 4,4,4Volume=a2h64=a2hT⋅S⋅A=2a2+4ah=2a2+4×64ad(T⋅S⋅A)da=4a−4×64a2a3=64a=4 h=4

The correct option is A 4,4,4
$V o l u m e = a^{2} h$
$64 = a^{2} h$
$T \cdot S \cdot A = 2 a^{2} + 4 a h$
$= 2 a^{2} + \frac{4 \times 64}{a}$
$\frac{d (T \cdot S \cdot A)}{d a} = 4 a - \frac{4 \times 64}{a^{2}}$
$a^{3} = 64$
$a = 4$ $h = 4$