Volume is given by:
.........[1]
Cost with regard to surface area is:
..........[2]
Solve [1] for y and sub into [2]:
Diferentiate, set to 0 and solve for x. y will follow.
Hey everyone, here is a minimization problem from the Optimization II chapter in my book. I have the maximization down, but this I can't seem to get. Can someone work through this one for me and show the steps so I can see the flow?
Minimizing Packagin Costs.
A rectangular box is to have a square base and a volume of 20 ft cubed. If the material for the base costs 30 cents/square foot, the material for the sides costs 10 cents/square foot, and the material for the top costs 20 cents/square foot, determine the dimensions of the box that can be constructed at minimum cost.
Hello, 2taall!
We use the same procedure of Minimization as for Maximization.
The tricky part (of course) is setting up the function to optimize.
I'll walk you thorough the set-up.
A rectangular box is to have a square base and a volume of 20 ft³.
The material for the base costs 30˘/ft², for the sides 10˘/ft², and the top 20˘/ft².
Determine the dimensions of the box that can be constructed at minimum cost.The volume of the box is: . .[1]Code:* - - - - * / / | / / | * - - - - * | | | | | | | y | | * | | / | | / x * - - - - * x
The base has area ft². .Its cost is: . cents.
The top has area ft². .Its cost is: . cents.
The sides have area ft². .Their cost is: . cents.
Hence, the total cost is: . .[2]
Substitute [1] into [2]: .
Hence, we must minimize: .
Ok guys,
I differentiated .50x^2 + 8/x
I got f'x= x - (8/ x^2 )=
solving for x:
-8/x^2 = -x =
-8= -x^3 =
8=x^3 =
taking the cubed root of each side:
x=2
Can you guys check that math?
Now we have(hopefully) x=2 and y= 20/x^2
EDIT: is the x=2 plugged back into the (x^2)y=20?
in which case [(2)^2]y=20 would make y=5
so the box would be 5 x 2 x 2 ?
Thanks everyone.
Thanks for looking it over!
Feel free to dabble in my:
Very hard area and definite integral problem
thread in the calculus forum too!