Find the cost of the cheapest container.

[Stuckonproblems]

A rectangular storage contrainer with an open top is to have a volume of 20m³. The length of its base is twice the width. Material for the base costs $20 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

 

[Downtrodden4]

Hope this helps

A rectangular storage contrainer with an open top is to have a volume of 20m³. The length of its base is twice the width. Material for the base costs $20 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

So in essence you are trying to optimize the space, to be the smallest amount of surface area while maintaining a constant volume. The use of a derivative would be great, and would result in optimization.
In order to do optimization, we require two formulas that can be proven by the statement above.

The two formulas should be as followed.
1. V=W*B*H
Volume=width*base*height
2. SA= 2(W*H)+2(B*H)+(B*W) NOTICE: The base*width is not multiplied by two because it has an open top

In optimization, the derivative is taken of the formula that is trying to be optimized. In this case Surface Area is trying to be optimized so you should take the derivative of the second equation. So plug in the variables we know and the conditions given above.

V=20 and 2W=B
So therefore we can turn the above Volume equation into this:
20=W*2W*H
Which can simplify to:
20=2W²*H
After some simple algebra
10/W²=H

Now that all the variables are taken care of the rest is just plugging in and a little calculus
Since W*H and B*H both equal $6 a square foot and B*W is $20 a square foot, you can also input this into the equation for surface area.

SA=2(W*H)+2(B*H)+(B*W)
Inserting the costs and inserting H=10/W² and 2W=B will change the equation to:
SA=240/W+1440/W+800W
SA=1680/W+800W
Then take the Derivative of this equation
SA'=-1680/W²+800
SA' is going to be a derivative of a constant which is always 0
0=-1680/W²+800
1680/W²=800
1/W²=800/1680
1/W²=2.1
W=.690
So bring the W to the equation before you took the derivative, in the 1680/.690+800*.690
The Final cost for the optimization should for the equation 2986.78

Hope this helps!

 

[Stuckonproblems]

Can you explain how you inserted the equations and costs to get the answer down here. I tried it but I didn't get it. This is what I get if I insert in the H=10/W² and 2W=B.
SA=2(W*10/W²)+2((2W)(10/W²))+(2W*W)
SA=2(10/W)+2(2(10/W))+(2W²)

Did I do something wrong on mine?

This is yours.
SA=2(W*H)+2(B*H)+(B*W)
Inserting the costs and inserting H=10/W² and 2W=B will change the equation to:
SA=240/W+1440/W+800W

 

[renegade05]

Can you explain how you inserted the equations and costs to get the answer down here. I tried it but I didn't get it. This is what I get if I insert in the H=10/W² and 2W=B.
SA=2(W*10/W²)+2((2W)(10/W²))+(2W*W)
SA=2(10/W)+2(2(10/W))+(2W²)

Did I do something wrong on mine?

This is yours.
SA=2(W*H)+2(B*H)+(B*W)
Inserting the costs and inserting H=10/W² and 2W=B will change the equation to:
SA=240/W+1440/W+800W

Yes, don't bother with S.A. Just go directly to the cost function. You are right, Downtrodden4 is inccorrect.

Your cost function is: \(\displaystyle C = 40w^2 + \frac{360}{w} \)

 

[Downtrodden4]

Sorry

WOW, didnt even think of the cost function. THAT would be much simpler than optimization of surface area, very sorry stuckonproblems