1. There is a parabolic arch bridge, as shown in the figure. At normal water level, the width of the water surface under the bridge is 20m, and the vault is 4m away from the water surface. (1) Find the expression of parabola in the rectangular coordinate system as shown in the figure; (2) On the basis of the normal water level, when the water level rises by h meters, the width of the water surface under the bridge is d meters, and the resolution function of h about d is found; (3) At normal water level, the water depth under the bridge is 2m. In order to ensure the smooth navigation of passing ships, the water surface width under the bridge should not be less than18m. When the water depth exceeds several meters, it will affect the smooth navigation of ships under the bridge.

Solution:

1. According to the topic, it is easy to get the quadratic function (0,0), (10, -4), (10, -4). Substituting these points into the general formula of quadratic function, we get the solution: a =- 1/25, b = 0. So the analytical formula is y =- 1/25x 2.

2. first use h to represent y? Y=-(4-h)？ And x=d/2? So substitute the above analytical formula.

-(4-h)=- 1/25(d/2)^2

3. The water surface width is 18, that is, x=9? Y side =-8 1/25? From bridge top to bridge bottom * * * 4+2 = 6m.

Therefore, when the water depth exceeds 6-8 1/25=2.76 meters, it will affect the smooth flow under the bridge.

2. It is known that a leasing company rents out 40 sets of equipment of the same model, and all of them are rented out when renting 270 yuan every month. On this basis, if the monthly rent per unit increases by 10 yuan, 1 unit of equipment will be rented out less. There is no rented equipment, and each set of 20 yuan needs to pay various fees every month. ?

Suppose the actual monthly rent of each set of equipment is X yuan (x≥270 yuan) and the monthly income is Y yuan (total income = equipment rental income-equipment cost not rented out)?

Question 1:? Find the quadratic function relationship between y and x?

Question 2:? When is the value of X and what is the maximum monthly income? What is the maximum value? ?

Question 3:? What is the monthly income when renting 300 yuan/apartment and 350 yuan/apartment? According to the calculation results of monthly income, how many sets of equipment should the company choose to lease at this time? Please briefly explain. ?

( 1)f(x)= x[40-(x-270)/ 10]-20 *(x-270)/ 10？

(2)f(x)=- 1/ 10x^2+65x+540？

f(x)=- 1/ 10(x-325)^2+ 1 1 102.5？

When x is 325, the monthly income reaches the maximum 1 1 102.5. ?

(3) Equal monthly income.

3. The cost of a commodity is 120 yuan, and the relationship between the product-free sales price x (yuan) and the product daily sales volume y (units) in the trial sales stage is Y=-X+200. In order to obtain the maximum sales profit, what should be the sales price of each product? What is the daily sales profit at this time?

Sales profit = (sales price-cost) * sales volume?

Sales price: x?

Sales volume: Y=-X+200?

Sales profit = (x-120) * y = (x-120) * (-x+200)?

=-X^2+200X+ 120X-24000？

=-X^2+320X-24000？

=-(X- 160)^2+ 1600？

Therefore, when the sales price x is 160 yuan, the sales profit is the largest, and the maximum is 1600 yuan.

This is quite classic!

I wish you progress in your study! ! !