10+x16-2x-2x^2 -4x +160 ≥ 130$12 or $13$3
Step-by-step explanation:
If x represents the number of $1 increases from $10 in the cost of the buffet, then the cost per customer is ...
c(x) = 10 +x
If n represents the number of customers for a given number of $1 increases (x), then we have ...
n(x) = 16 -2x
The revenue will be the product of these:
r(x) = c(x)n(x) = (10 +x)(16 -2x)
r(x) = -2x^2 -4x +160
Then the desired inequality is ...
r(x) ≥ 130
-2x^2 -4x +160 ≥ 130
The solution to this is ...
-2x^2 -4x +30 ≥ 0 . . . .subtract 130
x^2 +2x -15 ≤ 0 . . . . . . divide by -2
(x +5)(x -3) ≤ 0
The factors both have the same sign (hence a positive product) for x < -5 or x > 3. One of them is negative in the interval (-5, 3), so that is the solution to the inequality.
-5 ≤ x ≤ 3
Noah could charge $12 or $13 and maintain his revenue.
The maximum possible increase that maintains Noah's revenue is $3. This is the upper end of the solution space for the inequality.
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