Question

This is probably easy but i suck at word problems.

If 1,200 cm^2 of material is avaliable to make a box with a square base and a open top, find the largest possible volume of the box.

Answer 1

I have \[V = hwl\] and \[Sa = lw + 2wh + 2lh\]

Answer 2

But the base is a square so l = w

Answer 3

if you call the length of the base b and the height h you have
\[V=b^2h\] and also
\[b^2+4bh=1200\]solving the second for h give
\[h=\frac{1200-b^2}{4b}\] or
\[h=\frac{300}{b}-\frac{1}{4}b^2\]and therefore \[V(b)=b^2(\frac{300}{b}-\frac{1}{4}b^2)\] i believe

Answer 4

i just typed it in so maybe there is an algebra mistake but i think that is right. multiply out, take the derivative, find the critical point and that should do it

Answer 5

no i made an algebra mistake. should be
\[h=\frac{300}{b}-\frac{1}{4}b\]

Answer 6

\[V(b)=b^2(\frac{300}{b}-\frac{1}{4}b)\]

Answer 7

ahh ok.. thank you Satellite!!

Answer 8

\[V(b)=300b-\frac{1}{4}b^3\]
\[V'(b)=300-\frac{3}{4}b^2\] etc. good from there?

Answer 9

Yep!! Thank you!!

Answer 10

yw

Answer 11

So you have SA=1200=w^2+4wh or h = (1200-w^2)/4w
V=w^2h
Substituting: V=(1200-w^2)/4w*w^2 or V = [(1200-w^2)w]/4
V=(1200w-w^3)/4 or V=300w - w^3/4
V'=300-3/4w^2
There will be a maximum function value when the slope is 0 so solve 300-3/4w^2 = 0 and get w = 20. Since w = 20, l = 20 and h = (1200-w^2)/4w or (1200-400)/80 or 800/80 or 10. So the largest possible volume is 20 by 20 by 10 or 4000 cubic centimeters.