Question

a cuboid container with a base length twice its width is to be made with 48 metres squared of metal.
a) show that the height, h = 8/x - 2x/3, where x is the width of the base

b) express the volume, V, in terms of x

c) find the maximum volume

Answer 1

start by letting x = width so the length is 2x

the areas top and bottom = 4x^2 (width = x, length = 2x) 2 faces
front and back = 4xh (2x and height h) 2 faces
left and right = 2xh (x and height h) 2 faces

then using the information

\[48 = 2x^2 + 4xh + 2xh~~~or~~~~ 48 = 2x^2 + 6xh\]

then

\[48 - 2x^2 = 6xh\]

so

\[h = \frac{48 - 4x^2}{6x} \]

you can simplify it from here


so the volume is

\[V=lwh~~~or~~~ V = 2x \times x \times (\frac{8}{x} - \frac{2x}{3})\]

you can simplify this...

then the max volume

find the 1st derivative

and solve for x.

This will be the value that gives the max volume

hope it helps

Answer 2

thank you very much for the reply. I understand it but I'm just confused as to how you got from \[48 - 2x^2 = 6xh\] to \[\frac{ 48 - 4x^2 }{6x}\] where did the 4x come from?

Answer 3

oops the 1st line should be for surface area

48 = 4x^2 + 6xh

so

48 - 4x^2 = 6xh

Answer 4

okay thanks alot! :) I hope this doesn't come off as stupid but is that the answer or do I need to use the quadratic formula to find x.. Im still kind of confused what the question is asking

Answer 5

well the volume becomes

\[V = 2x^2 \times \frac{8}{x} - 2x^2 \times \frac{2x}{3}\]

so the volume equation is

\[x = 16x - \frac{2x^3}{3}\]

Answer 6

oops it starts with

\[V = 2x \times x \times (\frac{8}{x} - \frac{2x}{3})\]