# Algebra – Can’t figure out this word problem?

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Your furniture store sells two types of dining room tables. The first, type A, costs \$265 and you make a \$25 profit on each one. The second, type B, costs \$100 and you make a \$13 profit on each one. You can order no more than 40 tables this month, and you need to make at least \$760 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?

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I got:

A > 0

B > 0

A + B less than or equal to 40.

I forgot how to do this rest.

• let

A = number of type A tables

B = number of type B tables

Minimize

z = 265A + 100B

subject to:

A, B >= 1

25A + 13B >= 760

A + B <= 40

graph the lines, find the corner points and sub them into your minimize equation

(29.88, 1) = \$7,918.20

(39, 1) = \$10, 435.00

(20, 20) = \$7300

Cost is minimized at \$7,300.00 when 20 of each type is ordered

• 20 of each

profit 760

cost 7300

one way to solve the problem is to draw a graph with the X-axis being the number of type A and type B, assuming 40 altogether. the Y-axis is the profit. eg point 1 on the X-axis is 40xA, 0xB point 2 is 39xA, 1xB all the way to 1xA and 39xB. Calculate the profit for each combination and plot it, the target profit is 760

• To start setting up the equation – expenses go together on one side & profits on the other. The expenses are: \$4500 fixed overhead and \$50 for each product made. Profits: \$80 for each product made. x stands for the unknown number of products made. 4500 + 50x = 80x And you can continue from there. =) It’s nice to see that you aren’t just asking for the answer like so many others do!