Question

Problem 2 A company needs to order the material needed for constructing rectangular crates. The sides of these crates need to be a wire fencing material that can only be cut into whole number lengths and widths in order to properly assemble the crate. The crate is the shape of a right-rectangular prism. The sides of these crates are wire fencing that costs $5 or $6 per square foot (choose one). The top of these crates are made of hardwood that costs $20 per square foot. The bottom of these crates are made of a hard plastic that costs $10 per square foot. The crates need to be as low-cost as possible, but still retain a volume of 13 cubic feet. Also, the length of the crate must be twice the width Use calculus to discuss the "optimally" constructed crate. What are its dimensions?

Answer 1

Let the width of the box be ​​​​​​. Therefore we have length is ​​​​​, and the height is
2 3 2
​​​​​​. Cost of making the sides is
(2x 2) 195
​​​​​​. Cost of making the base and top is
202 40r260
. Therefore the total cost is
602.2. 195
​​​​​​. Therefore we want to minimize
c) 2195
​​​​​​. Since we know that the function is differentiable on all points other then 0, if the function has a minima, the derivative is 0 at that point. So we find the point at which the derivative is 0. Now
e'(r)-120r-195… (39) e'(z) = 120x
​​​​​​. Also from first derivative test we can see it is a point of minima. Once we have the width of the optimally constructed crate we can find the other dimensions. (technically we need to this entire process for 6 buck cost as well and we might get another point of minima, but whatever minimum cost we get, it can be further minimized by keeping the dimensions same and using the cheaper material).