Question

A manufacturer has determined that prot P (in dollars) is a quadratic function of dollars per month
a spent on advertising. The relationship is given by
p=1+50a-0.1a^2
if a is at most 500 dollars per month

-How much should be spent on advertising if prot is to be a maximum?
-How much should be spent on advertising if a prot of $5000 is desired?
-The manufacturer is spending $350 per month on advertising. Should the manufacturer increase
or decrease that amount? Explain.

Answer 1

a) where is the vertex?
.1 a^2 - 50 a -1 = -p

a^2 -500 a -10 = -10 p

a^2 - 500 a = -10 p + 10

a^2 - 500 a + 250^2 = -10 p + 62510

(a-250)^2 = -10 (p-6251
so
max profit of 6251 at a = 250

Answer 2

5000 = 1+ 50 a -.1 a^2

.1 a^2 - 50 a - 1 = -5000

.1 a^2 - 50 a + 4999 = 0
solve quadratic roots are 250+/-3sqrt1390
250 +/- 112
so
362 or 138

Answer 3

max profit is at 250 so if spending more than that, spend less