PRE-CALCULUS 11 Unit 1 – Day 9: OPTIMIZATION PROBLEMS

Optimization problems ask you to find the best amount and/or how to get that best amount. The best amount would be a maximum or minimum value. We will be solving optimization problems modelled by quadratic functions. What gives us the max or min y value? ______________________

Solving problems using algebra requires four items in the solution to earn full marks:

1) Define variables introduced and write expressions for all important quantities. One of the two variables defined must be the quantity that is being optimized.

2) Write an equation that models the situation described in the problem. Use the variables and expression from 1) to write a quadratic function for the quantity that is to be optimized.

3) Solve the equation. In this type of problem find the coordinates of the vertex to find the maximum or minimum value.

4) Answer the problem. Write an English sentence to answer what the problem is asking you to find.

example: 40 m of wire fencing will be used to make a rectangular plot that encloses a play area for your pet dog. A wall of your house is to serve as one side of the plot. What dimensions would give a maximum plot area?

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Solution: 40 m of wire fencing will be used to make a rectangular plot that encloses a play area for your pet dog. A wall of your house is to serve as one side of the plot. What dimensions would give a maximum plot area?

1) Define variables and write expressions. The area must be maximized. We need an area function that is quadratic. The area of a

rectangle is the product of its width and length. Write expressions for the dimensions of the plot.

Let A represent the plot's area in square metres.Let x represent the length of the plot's side perpendicular to the house's wall in metres.Then 40 2x represents the length of the plot's side parallel to the house's wall in metres.

2) Write the quadratic area function. A = x (40 2x)A = 40x 2x2 A = 2x2 + 40x

3) Solve the equation; write it in vertex form. A = 2[x2 20x ]A = 2[x2 20x + 100 100]A = 2[(x 10)2 100]A = 2(x 10)2 2(100)A = 2(x 10)2 + 200

The vertex is (10,200) and a < 0

4) Answer the problem. The maximum area is from the A-coordinate of the vertex, 200. For this area, the side length perpendicular to the house's wall must be from the x-coordinate

at the vertex, 10. For this x-coordinate, the length of the side parallel to the house's wall is calculated using

40 2x , so 40 2(10) or 20.

Answer: The width of the plot is 10 m and the length is 20 m.

house

plot

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exercise: A software company currently charges $600 for a program and is currently selling 400 units each month. It is estimated that sales would decrease by 10 units for each $20 increase in price. What should the company charge for each unit?

Consider: What does the company want to optimize?How is this quantity calculated? Try some numerical calculations.(Hint - let the second variable be the number of $20 price increases.)

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exercise: Find two numbers whose difference is 8 and whose product is a minimum.

exercise: The flight of a golf ball is modelled by the equation h(d) = 0.002d2 + 0.4d, where h(d) is the height (in metres) of the golf ball when it has travelled a horizontal distance d (in metres). Determine the maximum height reached by the ball and the horizontal distance the ball travelled when it reached this maximum height.