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GalinKa [24]
3 years ago
6

Your furniture store sells two types of dining room tables. The first, type A, costs $210 and you make a $21 profit on each one.

The second, type B, costs $180 and you make a $20 profit on each one. You can order no more than 130 tables this month, and you need to make at least $2640 profit on them. If you must order at least one of each type of table, how many of each type of table should you order if you want to minimize your cost? Need to know answer and how to graph answer
Mathematics
1 answer:
zimovet [89]3 years ago
7 0

<u>Answer</u>

In order to minimize total cost sell A type 40 numbers and B type 90 numbers

<u>Solution</u>

furniture store sells two types of dining room tables.  

The first, type A, costs $210 and you make a $21 profit on each one.  

The second, type B, costs $180 and you make a $20 profit on each one.  

We can order no more than 130 tables this month, and you need to make at least $2640 profit on them.

Let x be the number table  Type A sold, type B sold = 130-x

To find the value of x

21x +(130-x)20=2640

solving we get x=40

There are two possibilities

1)

A type sold =40 B type sold = 130 - 40 =90

total cost = 40 x 210 +  90 x 180 = 24600

2)

A type sold =90 B type sold = 130 - 90 =40

total cost = (90 x 210) + (40x180)=26100

In order to minimize total cost, choose first possibility

sell A type 40 numbers and B type 90 numbers

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Answer:

For A = 32/5 and B = 8 the system of equations will have infinitely many solutions.

Step-by-step explanation:

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The general form of linear equation in two variables is given by:

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In the given equations, after comparing with the general form

a_1 = 4\\b_1 = 5 \\c_1 = 10\\a_2 = A\\b_2 =B\\c_2 = 16

"In order for a system of equations to have infinity many solutions,

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} "

Putting the values we get

\frac{4}{A} = \frac{5}{B} = \frac{10}{16}\\\frac{4}{A} = \frac{5}{B} = \frac{5}{8}\\Now\\\frac{4}{A} = \frac{5}{8}\\\frac{A}{4} = \frac{8}{5}\\A = \frac{8}{5} * 4\\A = \frac{32}{5}\\And\\\frac{5}{B} = \frac{5}{8}\\\frac{B}{5} = \frac{8}{5}\\B = 8

Hence,

For A = 32/5 and B = 8 the system of equations will have infinitely many solutions.

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