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.|
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?
|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:
|6.||In order to calculate the x-value we fill in the y-value in the formula for the length:|
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?
|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 + 70480y = 40y + 33600y = 40y + 33600y–1
|5.||To calculate for which value the costs are minimum we differentiate and solve the differentiated formula to zero:
|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.