Question

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximized, what should the side length of the base be? What should the height of the box be? What is the maximal volume of the box? Your answers should be in terms of A.
FInd: Side length(cm), Height(cm), and Volume(cm)

Answer 1

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm ,   Height =  $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm  ,  Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$  cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

1. $y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V =  $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$  

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$