"When to Get the Maximum Profit" Quadratic Function PPT Courseware 2

Description

"When to Get the Maximum Profit" Quadratic Function PPT Courseware 2

learning target

1. Experience the process of exploring issues such as the maximum profit in T-shirt sales, and understand that the quadratic function is a mathematical model of a type of optimization problem.

2. Be able to analyze and express the quadratic functional relationship between variables in practical problems.

3. Be able to use the knowledge of quadratic functions to find the maximum (minimum) value of practical problems.

Self-study test

1. Determine the maximum value of the following quadratic function, and find out what is the maximum value when the independent variable is what value?

(1) y=x2-2x+3; (2)h=-5t2+15t+10

(3) s=-2/3t2+8t; (4)s=-1/2t2+18

2. A store purchases a batch of daily necessities with a unit price of 20 yuan. If they are sold at a unit price of 30 yuan, 400 units can be sold within half a month. According to sales experience, increasing the unit price will lead to a reduction in sales volume, that is, the sales unit price per If the sales price is increased by 1 yuan, the sales volume will be reduced by 20 units. When the selling price is increased by how much yuan, the maximum profit can be obtained within half a month?

Solution: Assume that when the selling price increases by x yuan, the profit obtained within half a month is y yuan. Then

y=(x+30-20)(400-20x)

=-20x2+200x+4000

=-20(x-5)2+4500

∴When x=5, the maximum value of y is =4500

Answer: When the selling price increases by 5 yuan, the maximum profit can be made within half a month of 4,500 yuan.

A store sells T-shirts. It is known that the unit price when purchased in batches is 2.5 yuan. According to market research, the sales volume and unit price satisfy the following relationship: within a period of time, when the unit price is 13.5 yuan, the sales volume is 500 pieces, and the unit price is 13.5 yuan per unit. If you reduce it by 1 yuan, you can sell 200 more pieces. Could you please help analyze, at what unit sales price can you make the most profit?

If the sales price is x yuan (x≤13.5 yuan), then

Sales volume can be expressed as: 500+200(13.5-x) pieces;

Sales can be expressed as: x[500+200(13.5-x)]yuan;

The profit obtained can be expressed as: (x-2.5)[500+200(13.5-x)]yuan;

When the sales unit price is 9.25 yuan, the maximum profit can be obtained, and the maximum profit is 9112.5 yuan.

Y=-200x2+3700x-8000

=-200(x2-18.5x)-8000

=-200(x2-18.5x+9.252-9.252)-8000

=-200(x-9.25)2+200×9.252-8000

=-200(x-9.25)2+9112.5

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Update Time: 2024-09-05

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