Middle School Mathematics Application of a function

When t=, q=40-5*

2.Set the talk time x cents, the cost y

Then y A = y B = when y A and y B are obtained, that is, when x<100, choose B;

When Y A is 100, choose A;

When yA=yB, that is, x=100, A and B are the same;

That is, y=-500x+60000

1.q=40-6t When t=, q=25l2Set the cost y, talk x minutes.

yA=yB=

When y-a-y-b, >, the solution is x<100

When y-a-y-b, >, the solution is x<100

When Y A 100

3.If there are x pieces of product A, then there are 50-x pieces of product B.

y=700x+1200(50-x)=-500x+60000

Solution: According to the topic. q=40-6t When t=, q=40-6*2Solution: Set the cost y, talk x minutes y a=

y B = when y A > y B, >, the solution is x<100

When y-a-y-b, >, the solution is x<100

When Y A 100

3.Solution: If there are X pieces of product A, then there are 50-x pieces of product B.

So y=700x+1200(50-x)=-500x+60000

<> as shown in the figure, the red line is Xiao Ming's itinerary, and the blue line is his mother's itinerary. The sum of Xiao Ming's reversible limb oak itinerary and empty return mother's itinerary is 2 kilometers of hunger. It took X hours for Xiao Ming to set off to meet his mother.

It took him 2 4 1 2 hours to get from home to school, X-1 2 hours back, and X-28 60 hours to get from home to school. The column equation is.

x-1 2)*6+(x-28 60)*5 2x=2 3 (hour).

Solution: From the question, it can be seen that Xiao Ming's home is 2 kilometers away from the school, and Xiao Yu Zhengming goes to school at a speed of 4 kilometers per hour, and it takes Xiao Ming 30 minutes to get to school.

It took X minutes for Xiao Ming to set off from the house to meet his mother, (x-30) minutes to return, and (x-28) minutes for Ma Hong to destroy his mother.

The formula is: 6(x-30) 60+5(x-28) 60=2 solution x=40 (minutes).

A: It took 40 minutes for Xiao Ming to travel from home to meet his mother.

I wish the landlord academic progress in the new year (*

It took Xiao Ming x minutes to set off from home to meet his mother, Xiao Ming took 2 60 4 = 30 minutes to arrive at school, and returned to Lao Pei (x-30) to serve Jingwei, and his mother used (x-28) points, so he got it.

6(x-30)/60+5(x-28)/60=2

x=40 (min).

From B to C to X Taiwan, the shipping fee is 3X

Then from B to D to transport (6-x) units, freight 5 (6-x) C needs 10 units, from A to C needs (10-x) units, freight 4 (10-x) D requires 8 units, from A to D requires 8-(6-x) = (2+x) units, freight 8 (2+x).

y=3x+5(6-x)+4(10-x)+8(2+x), i.e., y=2x+86

6-x≥010-x≥0

x≥0，2x+86≤95

Solution: 0 x

x is a positive integer, x, a common kind of transportation scheme.

y=2x+86

k=2>0, y increases with x, x when x=0, y min. = 86

(1) y=3x+(10-x)*4+(12-10+x)*8+(6-x)*5=2x+86 (10,000 yuan).

2）x=0 y=86; x=1 y=88; x=2 y=90; x=3 y=92;x=4 y=94 These 5 options are fine.

3) x=0 The minimum total cost is 860,000 yuan.

The cost of factory A = 600 + 2000 * yuan;

The cost of factory B = 600 + 2400*, so.

Suppose that when printing x (x 2000) copies, the cost of the two teams is equal, so.

600+2000*, the solution is x=4000, so.

When x 2000, the cost to A and B is the same;

When it is 2000 x 4000, you can get a bigger discount at factory A;

When x 4000, you can get a bigger discount to factory B.

Solution: Because car X is used in type A, car 50-x (1) y= is used in car type B

2) According to the theme to transport the goods, there is the following equation A: 35x+25(50-x)>=1530B: 15x+35(50-x)>=1150

The solution of the two equations is:

28<=x<=30

There are 3 kinds of schemes for arranging the number of carriages of A and B according to the requirements, 28 carriages of A and 22 carriages of B;

29 cars in A, 21 cars in B;

There are 30 cars in car A and 20 cars in car B.

3) y=The larger the value of x, the smaller, if you want to have the least freight, then y is the smallest, at this time, only when x is 30, the value is the smallest.

At this time, y = 10,000 yuan.

That's about it.

Take a look at it yourself.

(1) If B x is set, then C (8-x) is set.

x+ solution: x=2

So, two cars for B and six cars for C.

2) Set A a and B B, then C (20-a-b) 2A + B+

Solution: a=12-b......（1）

Let the profit be $y.

y=5a+7b+4(20-a-b)……2) Bring (1) into (2).

Get: y=92+2b

The bigger the b, the greater the profit, and the maximum of b is 18, so the maximum of y is 128

To divide the area of the triangle in two, l needs to pass through the line d(3,, which is x=3.

The coordinates of the point in the line ab (3,, so the equation is x=3