"When to Get the Maximum Profit" Quadratic Function PPT Courseware 4

"When to Get the Maximum Profit" Quadratic Function PPT Courseware 4

Apply what you learn: When to get the most profit

A store purchases a batch of daily necessities with a unit price of 20 yuan. If 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, every increase in the unit price of 1 Yuan, the sales volume is reduced by 20 pieces accordingly. Find the functional relationship between profit y and unit price x.

How to increase the selling price to get the maximum profit within half a month?

Solution: y=（x-20）[400-20(x-30)]

=-20x²+140x-20000

Do it

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 the price by 1 yuan, you can sell 200 more pieces.

Suppose 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.

Apply what you learn: As shown in the picture, Taohe Park will build a circular fountain. Install a pillar OA in the center of the pool perpendicular to the water surface. O is exactly in the center of the water surface, OA = 1.25m. From the nozzle at A at the top of the pillar outwards When spraying water, the water flow falls along a parabola with the same shape in all directions. In order to make the water flow shape more beautiful, it is required to design the water flow to reach a maximum height of 2.25m from the water surface at a distance of 1m from OA.

(1) If other factors are not considered, what is the minimum radius of the pool in m so that the sprayed water will not fall outside the pool?

Solution: (1) As shown in the figure, establish the coordinate system as shown in the figure. According to the question, the coordinates of point A are (0,1.25) and the coordinates of vertex B are (1,2.25).

Assuming that the parabola is y=a(x-h)²+k, the parabola expression can be obtained by the undetermined coefficient method: y=-(x-1)²+2.25.

When y=0, the coordinates of point C can be found to be (2.5,0); similarly, the coordinates of point D are (-2.5,0).

According to symmetry, if other factors are not considered, the radius of the pool must be at least 2.5m to prevent the sprayed water from falling outside the pool.

(2) If the parabolic shape of the water flow is the same as (1), and the radius of the pool is 3.5m, how many meters (accurate to 0.1m) should the maximum height of the water flow reach to prevent the water flow from falling outside the pool?

Solution: (2) As shown in the figure, according to the question, the coordinates of point A are (0,1.25) and the coordinates of point C are (3.5,0).

Assuming that the parabola is y=-(x-h)²+k, the parabola expression can be obtained by the undetermined coefficient method: y=-(x-11/7)²+729/196.

Or assuming that the parabola is y=-x²+bx+c, the parabola expression can be obtained by the undetermined coefficient method: y=-x²+22/7X+5/4.

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