# Differentiation and integration » Optimization

In optimization problems you are asked for the maximum or minimum value of a given variable. For example, you may be asked about the maximum size of the contents of a box or the minimum cost of the fences of a plot.

When solving algebraic optimization problems, follow the schedule underneath:
 1 Deduce from the question which two variables play a role. 2 Make a formula for both variables. 3 Rearrange the formula of the variable which is not asked. 4 Substitute the rearranged formula in the other. 5 Differentiate and solve the equation to zero. Because in the maximum or minimum the gradient is zero. 6 The found value must sometimes be filled in the other formula to find the other dimension or to calculate the minimum cost.

Example 1
A farmer wants a rectangular piece of land fenced with barbed wire. He also wants this piece of land divided into 4 equal pieces. See the drawing below. The farmer has a roll of barbed wire with a length of 360 metres. With which measurements will he get the maximum area?

Solution
1. The variables here are the length of the barbed wire and the area.
2. length = 2y + 5x and area = xy.
3. The length of the barbed wire is 360 metres.
The formula will become: 360 = 2y + 5x.
When you rearrange the formula to express x in terms of y you will get:
x = –0,4y + 72  (you subtract 2y and divide by 5)
4. When you substitute x in the other formula by this expression you will get:
area = (–0,4y + 72)y
Removing the brackets will give you: –0,4y2 + 72y.
5. In order to calculate the dimensions, where you will have the maximum area, this formula is to be differentiated and solved to zero:
 –0.8y + 72 = 0 –0.8y = –72 y = 90 metres
6. In order to calculate the x-value we fill in the y-value in the formula for the length:
 360 = 2 × 90 + 5x 360 = 180 + 5x 5x = 180 x = 36 metres

Example 2
A farmer is fencing off a rectangular piece of land. He also wants this land divided into 4 equal pieces. The fence at the edges of his land he uses a fence of 20 euros per metre. The other pieces cost 10 euros per metre fence. See the drawing below. The farmer wants a pice of land measuring 480 square metres. With wich measurements can he minimalize his costs?

Solution
1. The variables are area and costs.
2. area = xy and costs = 20y + 20y + 20x + 20x + 10x + 10x + 10x = 40y + 70x.
3. The area is 480 metres, filling this in will give you: 480 = xy. Rearranging gives you: x = 480y .
4. Substituting this formula for x gives you:
costs = 40y + 70 480y = 40y + 33600y = 40y + 33600y–1
5. To calculate for which value the costs are minimum we differentiate and solve the differentiated formula to zero:
 40 – 33600y2 = 0 33600y2 = 40 y2 = 840 y ≈ 29,0
6. If you fill in this y-value in the formula of the area, the x-value can be calculated.
x = 48029.0 ≈ 16.6 metres.