aanpalendmw

2022-07-18

Finding dimensions of a rectangular box to optimize volume
A rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of $16f{t}^{2}$. Compute the dimensions of the box that will maximize its volume.
I am going wrong somewhere but I can't see where. I have set $2\left(3h+3w+hw\right)=16$ and my optimization equation is $v=hwl$. I set $w=\frac{8-3h}{3+h}$ and plug into the optimization equation. Calculating v' gives me $\frac{9{h}^{2}-84h+72}{{\left(3+h\right)}^{2}}$,but setting that equal to zero does not give me the answer. I need for h, which is 1.

Abbigail Vaughn

Explanation:
The equation is going wrong you forgot to subtract the top so itll be $2\left(3b+bh+3h\right)-3b=16$ here b=breadth and h=height . Now you proceed by the way as you did and you will get an equation in h for volume diffrentiate it and put it to be 0. Thus we get eqn $\frac{48h-18{h}^{2}}{3+2h}$ so on differentiating wrt h we get a quadratic as ${h}^{2}+2h-4=0$ thus $h=\frac{-1+,-\sqrt{4+16}}{2}$ so we take positive sign for max area thus its $-1+\sqrt{5}$ now you can get breadth and you are done with the sum.

comAttitRize8

Step 1
Did you forget that the top of the box is open? The surface area is $2\left(3h+hw\right)+3w=16,$ so that $w=\frac{16-6h}{3+2h}.$.
Step 2
The volume of the box is $V=3hw=3h\frac{16-6h}{3+2h}.$
If you differentiate this with respect to h, and set it equal to 0, you get the h for which the volume is maximized.

Do you have a similar question?