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
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.
3 answers
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
.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
max profit is at 250 so if spending more than that, spend less
1 answer