Answer:
A: cost = 10+x; customers = 16-2x
B: -2x² -4x +160 ≥ 130
Step-by-step explanation:
In these price and revenue optimization problems, the relation between price and sales is given in the problem statement. After that is translated to math terms, the revenue as the product of price and sales can be determined.
<h3>Part A</h3>The variable x is defined as the number of $1 increases from a price of $10. That means ...
Cost = 10 +x
For each $1 increase, the number of customers decreases by 2 from a starting value of 16:
Customers = 16 -2x
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<h3>Part B</h3>The problem statement tells you the revenue (per hour) is the product of cost and number of customers. The owner wants this to be at least 130:
(Cost)(Customers) ≥ 130
(10 +x)(16 -2x) ≥ 130
160 -20x +16x -2x² ≥ 130 . . . . eliminate parentheses
-2x² -4x +160 ≥ 130 . . . . . . . . . put in required standard form
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<em>Additional comment</em>
The attached graph shows the solution to this quadratic inequality is ...
-5 ≤ x ≤ 3
and that revenue can be increased by $2 per hour by decreasing cost $1 (x=-1).
