We are going to fence in a rectangular field that encloses 200m ^2.

Question

If the cost of the material for of one pair of parallel sides is $ 3/m and cost of the material for the other pair of parallel sides is 8/m determine the dimensions of the field that will minimize the cost to build the fence around the field.

Answer 1

Let x = width and y = length dimensions of the rectangle

Area, A = xy = 200 m² --> y = 200/x

Cost, C = 3(2x) + 8(2y)

C(x) = 6x + 3200/x

Take the derivative and then set that to zero to solve for x

C'(x) = 6 - 3200/x²

x = √(3200/6) ≈ 23.094

y = 200/x ≈ 8.66

The width ≈23.094 m and the length ≈ 8.66 m.

C(23.094) ≈ $277.13

Answer 2

This is a super common set-up for an optimization question, and our game plan will be as follows:

Write an equation for the quantity we are optimizing (in this case, the cost). Frequently, this equation will have two different independent variables (in this case x and y, the dimensions of rectangle).

Use the constraint (in this case the area to be enclosed) to write an equation relating the two independent variables. Then solver for one of them, and substitute this expression in for the variable in the other equation.

Now we will have an equation for the quantity to be optimized that only has one independent variable, so we can take the derivative, set it = 0, and solve to find the ideal values. This is what it looks like:

x: less expensive side length (in m). y: more expensive side length (in m). C: cost (in $)

C = 6x + 16y (we have 2 sides of each length to build). A = xy = 200 so y = 200/x

Substituting we get C(x) = 6x + 3,200/x

C^'(x) = 6 - 3,200/x² = 0

6x² = 3,200

x = 40/√3 and y = 5√3

Notice their product is 200, and that we made the more expensive sides significantly shorter (which makes sense since we are trying to minimize cost).