# Marginal Profit Question

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• Jun 24th 2010, 08:34 PM
daftmau5
Marginal Profit Question
Im having a lot of trouble trying to figure out how to solve this problem, any help is greatly appreciated! Thanks.

The profit (measured in dollars) of a small company has been determined to be
P(x)= 20000 (x / 100 + x^2), where x is the production level (in units).

a) Compute the marginal profit

b) Find the production level xmax that leads to the maximal profit within the range
0 ≤ x ≤ 30.

c) How many dollars of profit does the company make at that production level?

d) Does your answer change if x has to be within the range 0 ≤ x ≤ 10 instead? If yes, how?
• Jun 25th 2010, 01:40 AM
SpringFan25
a)
the marginal profit is $\frac{dP}{dx}$

b)
maximal profit occurs either at the point where marginal profit = 0, or x=0, or x=30.

Find the profit at each of these points and see which one is higher
hintThere is no point in the range 0 < x < 30 where marginal profit =0 in this case.

c)
this is equal to P(x), where x is the production level you found in part b

d)
if your existing answer is in the range 0 < x< 10, then it will not change.

However if your existing answer is not in the range, you need to find a new one, which will be either:
x=0
x=10
or somewhere in between where the marginal profit is zero (you can discount this possibility as if there was a point in the range with marginal profit =0, you would have found it in part b)
• Jun 25th 2010, 09:07 AM
daftmau5
Thanks so much for your help!

Please let me know if these answers are correct:
A.)20000 [(100-x^2)/(100+x^2)^2]
B.)x=10
c)p=1000
d.) No, because in the previous problem, 10 had the highest amount of profit.
• Jun 25th 2010, 11:31 AM
SpringFan25
hmm, if i understood your function correctly i dont think you have the marginal profit right

You wrote P(x)= 20000 (x / 100 + x^2)

which is $P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2$

Did you mean

$P(x) = 20000 \left(\frac{x}{100 + x^2} \right)$?

I'll assume you meant $P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2$
a)
$P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2$
$MP = \frac{dP}{dx} = 200 + 40000x$

b) MP = 0 has no solutions where x > 0
So max pforit occurs at either x=0 or x=30
p(0) = 0
p(30) = 1200200

so x=30 maximises profit

c)
p(30) = 1200200

d)
Yes, the new maximum is at x=10