Question

a box has a square base of side x and height h find the dimensions x, h for which the volume is 14 and the surface area is as small as possible

Answer 1

Volume of cuboid :

$14 = x^2h\\\\h = \dfrac{14}{x^2}$

Surface area of this cuboid :

$A = 2( x^2 + xh + xh )\\\\A = 2( x^2 + 2xh)$

Putting $h = \dfrac{14}{x^2}$ in above equation, we get :

$A=2(x^2 + 2x\times \dfrac{14}{x^2})\\\\A = 2( x^2 + \dfrac{28}{x} )$

For minimum area, $\dfrac{dA}{dx}=0$ :

$\dfrac{dA}{dx}=2( 2x -\dfrac{28}{x^2})\\\\\dfrac{dA}{dx }= 4(x - \dfrac{14}{x^2})\\\\x - \dfrac{14}{x^2}=0\\\\x=\sqrt[3] {14} = 2.41$

Therefore, value of x is 2.14 .

Hence, this is the required solution.

Answer 2

The dimension for which the volume is 14 and the surface area is as small as possible are 2.41 by 2.41 by 2.41

The formula for calculating the volume of the square box is expressed as:

$V = x^2h$

Given that the volume of the box is 14, on substituting the value we will have:

$14 =x^2h\\h=\frac{14}{x^2}$

The surface area of the box is expressed as:

$S = 2(x^2+xh+xh)\\S=2(x^2+2xh)\\S = 2(x^2+2x(\frac{14}{x^2} ))\\S = 2(x^2+\frac{28}{x} )\\S=2x^2+\frac{56}{x}$

To get the surface area is as small as possible, then dS/dx = 0

$\frac{dS}{dx} = 4x-\frac{56}{x^2}\\0 = 4x-\frac{56}{x^2}\\4x = \frac{56}{x^2}\\4x^3 = 56\\ x^3=14\\x=\sqrt[3]{14}\\x= 2.41$

Get the height of the box. Recall that;

$h=\frac{14}{x^2}\\h=\frac{14}{(2.41)^2}\\h=\frac{14}{5.8081}\\h= 2.41$

Hence the dimension for which the volume is 14 and the surface area is as small as possible are 2.41 by 2.41 by 2.41

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