Math problems. Help If a merchant sells a commodity with a purchase price of 8 yuan at 10 yuan per

Solution: The set price is x, and the profit is yRule.

y=[100-10（x-10)](x-8)(200-10x)(x-8)

Finally, the quadratic function (the square of x is written as x2, and so on) y=-10x2+280x-1600

Find the maximum value of the function, and get y=-10(x-14)2+360 Therefore, when the price is 14 yuan, the profit obtained is the largest, and the maximum profit is 360 yuan.

Set a price increase of x yuan.

Now the single profit is 10 + x-8 yuan.

100-x can be sold, and the total profit is (10+x-8)*(100-x) yuan.

It's a quadratic function, just find the maximum value, don't forget to take the value range.

I'm a person who likes to use elementary school arithmetic methods to solve problems.

According to 10 yuan**, the profit is:

10 8) 100 200 (yuan).

If you press $11**, the profit is.

11 8) (100 10) 270 (yuan) Since you want to be the maximum, then the two factors should be centered as much as possible.

So it should be the profit when the price is 14 yuan.

14 8) (100 40) 360 (yuan).

Set the selling price at x

Sales quantity: 100-10 (x-10).

Then let the profit be y

y=（x-8）[00（x-10）]

Then it can be concluded that this is a quadratic function, which can be found after the formula.

If the set price is x, then the profit n = (x-8) * [100-10 * (x-10)] = -10x 2 + 280x-1600

It is not difficult to conclude that the profit is the largest when x = 14.

This is a maximum-value problem for a one-dimensional quadratic equation. Column equations can be answered.

Solution: The set price is X yuan, and the sales volume is 200-10 (x-10).

Profit y=(x-8) (200-10 (x-10) 20(x-8)(20-x)=-20(x2-28x+160)=-20(x-14)2

720, when x = 14, the y value is the largest, which is 720, therefore, when the selling price is set at 14 yuan, the profit is the largest, and the maximum profit is 720 yuan.

10 yuan, 8 yuan, 2 yuan.

2 yuan 200 400 yuan (this is the money that should have been made before the price increase) (2 400

Where: 2 is 2 yuan, x is the number of price increases, is how many yuan has risen, 200 is 200 pieces, 10x is every time the price rises for yuan, it will reduce the sales of 10 pieces.

400 Yes, the purpose of the price increase must be greater than the profit of 400 yuan before the price increase.

2 is the profit earned for each item after the price increase, (200 10x) is how many items are sold today.

2 is today's profit.

Solution (2 400 20x 100x 5x 400.)

80x－5x²＞0

x²－16x ＜0

x 16, so take x 15

Profit is greatest when you earn $2 per item.

The maximum profit is ($2).

The price is 8 yuan).

The set price is $x, and the sales volume is 200-10 (x-8).

Maximum profit y=x (200-10 (x-8)).

Let each price increase be x yuan (x 0), and the profit will be y yuan;

Daily sales (100-10x);

The total sales per day is (10+x) (100-10x) yuan;

The total purchase amount is 8 (100-10x) yuan

Apparently 100-10x 0, x 10

y=（10+x）（100-10x）-8（100-10x）=-10x2+80x+200

10 (x-14) 2 + 360 (0 x 10), when x = 14, y obtains the maximum value of 360, so the sales unit price is 14 yuan, and the maximum profit is 360 yuan

Solution: Set the price increase of x yuan, that is, the selling price is (10 + x) yuan, and the profit is y yuan

y=（10+x-8）（100-10x）

2+x）（100-10x）

10x²+80x+200

y= -10（x-4）²+360

When the price increase x is 4 yuan, that is, the selling price is 10 + 4 = 14 yuan, the profit earned per day is the largest, and the maximum profit is 360 yuan.

Using the step-by-step test method, calculate the profit when the selling price is 11 yuan, 12 yuan, 13 yuan, 14 yuan, and 15 yuan.

Solution: Let the selling price be positioned at X yuan, so that the profit can be maximized, and the profit can be yy=(10+ (x>=0).

Simplification: y=-50x+80x+400(x>=0) Because -50x<0 so there is a maximum parabolic opening downward, so when x=, y is the largest, that is, the profit earned is the largest, y=yuan.

Solution: When the selling price of this commodity is reduced by X yuan, the profit is the largest, and the maximum profit is y yuan, then y = (10-8-x) (100 + 100, so when x = -b 2a = -100 2 (-100) = yuan, the maximum profit is y = 4ac-b2 4a = 225 (yuan).

Answer: Reducing the selling price of this commodity can maximize the sales profit, and the maximum value is 225 yuan

If you reduce the unit price by $x, the unit price is (10-x).

Profit y=(10-x-8).

To simplify, y=100(-x*x+x+2)=100[-(x-1 2)squared +9 4].

So the profit is maximized when x=. Maximum profit = 100 * 9 4 = 225

The maximum profit is 225 yuan when the price is reduced.

Equation: Let be x, and the profit will be y

y=[(10-x)-8]×（100＋100x）

(1) y=-10x2 +280x-1600 (2) 14 yuan test question analysis: (1) according to the average relationship of the question: profit = (selling price - purchase price) The number of pieces sold, the profit per piece is (x-8) yuan, because 10 yuan per piece is sold 100 pieces, and every 1 yuan is increased, the number of pieces is 10 pieces less, then the number of pieces is 100-10 (x-10) pieces, and the equation is:

y=(x-8)[100-10(x-10)], i.e., y=-10x2 +280x-1600;

2) The opening of the function is downward, and the profit is the highest, which is to find the ordinate of the vertices of the function, and the equation formula in (1) obtains:

y=-10(x-14)2 +360, when x=14, y max = 360 yuan, answer: when the selling price is 14 yuan, the profit is the largest.

Comments: This question is a common exam question, which mainly examines students' application of quadratic functions in practice, first analyzes and clarifies the relationship between x and y, and then lists the relationship between functions, and finds the maximum value through the properties of functions.

10*100=1000

x*(20-x)*10=y

There is no restriction on the purchase price.

I'm sorry, I'm wrong, the profit is y, I'll change it for you.

x-8)*(20-x)*10=y

x=8 y=0

x=9 y=110

x=10 y=200

x=11 y=270

x=12 y=320

x=13 y=350

x=14 y=360

x=15 y=350

x=16 y=320

x=17 y=270

x=18 y=200

x=19 y=110

x=20 y=0

You see, can you?

For each reduction of yuan, its sales volume can increase by 10 pieces, reduce x yuan, its sales volume can increase by 100x pieces, the original profit of each piece is 10-8, now reduce x yuan, now the profit of each piece is 2-x, 2-x 0 should be guaranteed, sales profit y=(10-8-x) (100+100x)=-100x2+100x+200(0 x 2).

If you don't understand, you can ask and I'll tell you.