# Algebra II

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A furniture company displays bedroom sets which require 21 square meters of space and living room sets which require 42 square meters of space. The company which has 546 square meters of available space,wants to display at least 6 bedroom sets and at least 5 living room sets.

My question is:If a bedroom set sells for \$10,000 and a living room set sells for \$18,000,determine the number of bedroom sets and living room sets that must be sold to maximize the amount collected.
I'm not sure what I need to do to set this up.

• Algebra II -

Think it through.. don't try to find a single magic equation.

You maximize receipts when you sell as many living room sets as you can, while still being able to display and sell the required minimum 6 bedroom sets. Since both take up 21 sq meters and you have 546 sq m available, you can display 546/21 = 26 sets of either kind. So, display the required 6 minimum bedroom sets and use the rest of the space for 20 living room sets. Assuming that they all sell at an equal rate (which is not at all likely in practice), the amount collected after one turnover will be 20*\$18,000 + 6*\$10,000 = \$420,000

I did not realize furnishing a house is that expensive nowadays

• Algebra II -

Let the number of bedroom setups be b
then the number of living room setups is (546-21b)/42

total revenue
= 10000b + 18000(546-21b)/42
= 234000 + 1000b

but b ≥ 6 and livingrooms ≥ 5

so if b=6 , l= 10
and if l=5, b = 16

the total revenue is a linear function represented by a straight line with a positive slope, so the larger value of b will give a larger value of revenue

so when b= 16, l = 5 and revenue = \$250,000

check: if b = 16, l = 5 revenue = \$240,000

So they should set up 16 bedrooms and 5 living rooms

• Algebra II -

Correction:
Second last line should say:

check: if b = 6, l = 10 revenue = \$240,000

• Algebra II -

I misread the problem, not noticing the higher area of the living room sets. Just ignore my answer, if you haven't already, and follow Reiny's method.

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