Question

A rectangular box whose volume is 32 is open at the top. If the surface of the area is 2( L + B )H +LB, where L L, B, H are length, breath and height respectively.

(A) Find the dimension of the box that may regure least material

(b) Investigate weather the dimension found require least material.

Answer 1

Solution;

Let the base of the box be square,such that,

$L=B$

Volume of the box is;

$V=L^2H$

By substitution;

$32=L^2H$

Make H the subject of the formula;

$H=\frac{32}{L^2}$

The surface are will be;

$S.A=L^2+4LH$

Substitute for H;

$S.A=L^2+\frac{128}{L}$

For least material, minimise the Surface Area;

$S.A'=2L-\frac{128}{L^2}=0$

$L^3=64$ ; $L=4$

Therefore the for material;

The base should have a square base of 4×4 and a height of H=2

(b)

Using the second derivative;

$S.A''=2+\frac{256}{x^3}$

The value is positive at L=4