An open on the top box whose length is twice as long as the width has the volume of 600 in. The material for the bottom base costs $3 per square inch, the material for the sides cost $2 per square in.

Question

I need to find the minimum total cost. I know it occurs between a width of 6-8. I am struggling trying to figure out the primary and secondary equations needed to solve this problem. The cost function, and the volume.

I tried the cost function would be (h(fx))= 3(2x^2)+2(2xh+4xh)

and the secondary equation being 600= 2x^2+2xh+4xh

but it ends up not working out after I solve the secondary equation for h which i got h=(300-x^2)/3x

then substitute that for h in the primary equation.

I have no clue what all I have that is wrong and could just really use some help!

Thank you!

Answer 1

LWH=600

L=2W => H=600/2W²

Cost C = 3LW + 2(2LH + 2WH)

Use the first two equations to write C as a function of W alone, then get dC/dW and set it equal to 0; then solve for W. It's messy, but not difficult.