If 1200 cm^2 of material is available to make a box

Solution:

Let “b” represent the length of the base and “h” represent the height. The surface area of the box can be represented as

$1200=b^{2}+4hb$

and isolating for “h” gives us,

$h=\frac{(1200-b^{2})}{(4b)}$

The volume is $V=b^{2}h=\frac{b^{2}(1200-b^{2})}{(4b)}=300b-\frac{b^{3}}{4}$

Thus,

$V'(b)=300-\frac{3}{4}b^{2}$

Now, we see where the absolute maximum is for our function by equating $V'(b)=0$.

$0=300-\frac{3}{4}b^{2}$

$300=\frac{3}{4}b^{2}$

$b^{2}=400$

$b=20$.

Therefore, the absolute maximum is at b=20. Substitute this value back into $h=\frac{(1200-b^{2})}{(4b)}$

$h=\frac{(1200-20^{2})}{(420)}$

$h=10$

Thus, the largest possible volume is $b^{2}h=(20)^{2}(10)=4000cm^{3}$