I. Analysis of quantitative relations
According to the actual situation, the relationship between a certain number and quantity is given, and it is required to write the functional relationship directly and apply it on the basis of analysis.
The key to solve this kind of quadratic function application problem is to carefully analyze the meaning of the problem and write out the quantitative relationship correctly.
Example 1. A chemical raw material distribution company purchases 7000 kilograms of a chemical raw material, and the purchase price is 30 yuan per kilogram. The price department stipulates that its sales unit price shall not be higher than that of 70 yuan per kilogram, nor lower than that of 30 yuan. The market survey found that when the unit price is set at 70 yuan, the average daily sales volume is 60 kg; Every time the unit price is reduced by 1 yuan, it will sell 2 kilograms more every day. During the sales process, other expenses will be paid every day (500 yuan) (if the number of days is less than one day, it will be counted as a whole day). Suppose the sales unit price is X yuan and the daily average profit is Y yuan.
(1) Find the quadratic function relation of y about x and indicate the value range of x;
(2) Formulating the quadratic function obtained in (1) and writing the vertex coordinates; Draw a sketch in the coordinate system shown in Figure 2; Look at the pictures and point out how much unit price can be set every day to get the most. What is this?
(3) If all the chemical raw materials are sold, compare the two sales methods, the most average daily profit and the highest sales unit price, which one gets more total profit and how much more?
Analysis: (1) If the sales unit price is X yuan, it will be reduced by (70-x) yuan per kilogram, the average daily sales will be [60+2(70-x)] kg, and the profit per kilogram will be (x-30) yuan. According to the meaning of the question (30≤x≤70).
(2)。 The vertex coordinate is (65, 1950), and the sketch is omitted. When the unit price is set to 65 yuan, the daily average profit is the largest, which is 1950 yuan.
(3) Through formula calculation, when the average profit of the day is the highest, the total profit can be195,000 yuan; When the sales unit price is the highest, the total profit can reach 22 1500 yuan. Therefore, when the sales unit price is the highest, the total profit is more, and the extra profit is 221500-195000 = 26500 yuan.
Second, the undetermined coefficient method
The title clearly gives the quadratic function relationship between two variables and several pairs of variable values, and requires to find out the function relationship and make a simple application.
The key to solve this kind of quadratic function application problem is to use the undetermined coefficient method skillfully and find out the function relationship accurately.
Example 1. The products produced by a company cost 2 yuan and cost 3 yuan, and the annual sales volume is 1 10,000 pieces. In order to get better benefits, the company is going to spend some money on advertising. According to experience, when the annual advertising fee is X (10/00000 yuan), the annual sales volume of the product will be Y times of the original sales volume, and Y is the quadratic function of X. Their relationship is as follows:
X (one hundred thousand yuan)
1
2
…
y
1
1.5
1.8
…
(1) Find the functional relationship between y and x;
(2) If profit is regarded as total sales minus cost and advertising fee, try to write the functional relationship between annual profit S (100,000 yuan) and advertising fee X (100,000 yuan);
(3) If the annual advertising fee is 65,438+0-300,000 yuan, what is the range of the advertising fee, and the annual profit of the company will increase with the increase of the advertising fee?
Analytical solution: (1) Because the quadratic function relation of Y is X is given in the question, the function relation of Y and X can be obtained by using the undetermined coefficient method.
(2) S= 10y(3-2)-x from the meaning of the question.
(3) According to the properties of (2) and quadratic function, when 1≤x≤2.5, that is, the advertising fee is between10-250,000 yuan, S will increase with the increase of advertising fee.
Third, establish a mathematical model.
It is required to construct quadratic function independently and solve practical problems by using the images and properties of quadratic function.
This kind of quadratic function application problem has high requirements and certain difficulty. Example 3 .. An environmental protection equipment company sells a new product with great market demand. The purchase price of each product is known in 40 yuan, and the sales volume is calculated in the distribution process. Y (10,000 pieces) and the sales unit price X (yuan) have a linear functional relationship as shown in the figure, and the annual total expenditure Z (10,000 yuan) (excluding the purchase price) has a functional relationship with the annual sales volume Y (10,000 pieces).
(1) Find the functional relationship between y and x;
(2) Write the functional relationship between the annual profit w (ten thousand yuan) of the company selling the product and the sales unit price x (yuan); (Annual profit = total annual sales-total annual purchase price of products-total annual expenditure) When the sales unit price is X, what is the maximum annual profit? What is the maximum value?
(3) If the company wants the annual sales profit of this product to be no less than 575,000 yuan, please use the function image in (2) to help the company determine the range of the unit sales price of this product.
In order to maximize the sales of products under such conditions, what do you think the unit sales price should be?
Solution: (1) From the meaning of the question, let y=kx+b, and the image passes through (70,5) and (90,3).
The solution is y = x+12.
(2) From the meaning of the question, w = y (x-40)-z = y (x-40)-(10y+42.5) = (x+12) (x-10)-/kl.
=-0. 1x 2+ 17x-642.5 =(x-85)2+80。
When I was in 85 yuan, the highest annual profit was 800,000.
(3) Let w=57.5 and get-0.1x 2+17x-642.5 = 57.2.
After sorting, x2- 170x+7000=0.
The solution is x 1=70, x2= 100.
As can be seen from the image, if the annual profit is not less than 575,000 yuan, the sales unit price should be between 70 yuan and 100 yuan. And because the lower the sales unit price, the greater the sales volume. In order to maximize the sales volume and the annual profit is not less than 575,000 yuan, the sales unit price should be set as 70 yuan.