Question

A closed box is constructed of 4200cm2 of cardboard. The box is a cuboid, with height hcm and square base of side xcm . What is the value of x which maximises the volume of the box? Give your answer in cm correct to 3 significant figures. [Answer format: 00.0 cm]

Answer 1

Answer:

x is 10$\sqrt{7}$

Step-by-step explanation:

• Let x is the side of the base
• Let h is the height

Given that:

• The area A = 4200 cm2

<=> 2$x^{2}$ + 4xh = 4200 cm2

<=> 4xh = 4200 - 2$x^{2}$  

<=> h = (4200 - 2$x^{2}$  )/4x

• The volume of the box:

V = $x^{2}$  h

<=> V = $x^{2}$  (4200 - 2$x^{2}$  )/4x

<=> V = (4200x - 2$x^{3}$ )/4

• To find the maximum volume, we differentiate volume with respect to x:

dV/dx = (4200 - 6$x^{2}$ )/4

Set dV/dx = 0, we have:

4200 - 6$x^{2}$ =0

<=> $x^{2}$  = 700

<=> x = 10$\sqrt{7}$

• We differentiate again

d²V/dx² =  -12x/4

It is negative so the volume is maximum.

So x is 10$\sqrt{7}$