The design of a digital camera box maximizes the volume while keeping the sum of the dimensions at 6.75 inches. If the length must be 1.5 times the height, what should each dimension be?

Question

whpalmer4 · Answer

Okay, we have 
$$V = l*w*h$$$$l=\frac{3}{2}h$$$$l+w+h = 6\frac{3}{4} = \frac{27}{4}$$

What class is this for?  I don't want to use any techniques you may not have learned...

whpalmer4 · Answer

First, let's substitute for $l$ in the constraint equation:$$\frac{3}{2}h+w+h = \frac{27}{4}$$Solve that for $w$ in terms of $h$
$$w = \frac{27}{4}-\frac{5}{2}h$$
Now our volume can be written as $$V = l*w*h = (\frac{3}{2}h)*(\frac{27}{4}-\frac{5}{2}h)*h= \frac{81}{8}h^2 - \frac{15}{4}h^3$$
Do you have any idea how to find the value of $h$ that will maximize (or minimize) $V$?

anonymous · Answer

It's for algebra II

whpalmer4 · Answer

Great.  How about my other question?  Do you have any idea how to find the value of $h$ that will maximize $V$?  Unfortunately, "Algebra II" is not a standardized description of what you may have been taught :-)