I've been stuck on this problem forever. Can someone please explain to me what to do from here?
The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches. If the length must be 1.5 times the height, what should each dimension be?
I have:
v=lwh
l=1.5h
w=6-2.5h
1.5h+h+w=6
2.5h+w=6 (Then I solved for w)
v=(1.5h)(h)(6-2.5h)
Simplified to:
v=9h^2-3.75h^3
I don't understand what to do from here. Please help
Well, in calculus, it is easy, to give you a taste..
dV/dh=18h-3.75*3h+2=-
which has solutions at h=0, h=18/(3.75*3)
at h=0, the volume is minimum, and at the other h, it is maximum.
Method 2: Graph the equation 9H^2-3.75H^3 on your graphic calculator.