Question

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 time the height, what should each dimension be?

Answer 1

`v=1.5l*l*w`

`w=6-l-1.5l`

`v=1.5l*l*(6-l-1.5l)`
which simplifies to

`v=9l^2-3.75l^3`
and so we take the derivative of that function in terms of l

`(d)/(dx)v=18l-11.25l^2`
then we set that to 0

`-11.25l^2+18l=0`
using the quadratic formula

`(-b+or- √(b^2-4ac))/(2a)`

`(-18+or- √(18^2-4*11.25*0))/(2*-11.25)`
simplifying..

`(-18+or- √(324))/(-22.5)`

`(-18+√(324))/(-22.5)=0`

`(-18-√(324))/(-22.5)=1.6`

at this point i noticed an error, i used l instead of h... but thats ok

pluggin 0 and 1.6 for h, lets check answers
1.6: h=1.6 l=2.4 w=2
0: h=0 l=0 w=6

so the answer is h=1.6 l=2.4 w=2, which gives the greatest volume of 7.68