# Elasticity question: Finding maximized price.

• Apr 20th 2009, 10:50 PM
penbralat
Elasticity question: Finding maximized price.
The consumer demand curve for cases of gadgets is given by \$\displaystyle q = (90 - p)^2\$, where p is the price per case and q is the demand in weekly sales.

a) If the gadgets cost \$30 per case to produce, use calculus to determine the price that should be charged per case to maximize the weekly profit.

I know how to find revenue but the \$30 in the equation's throwing me off. On my answer key the answer is \$50 per case but I keep coming up with \$45. Here's what I am doing:

\$\displaystyle R=pq\$
\$\displaystyle p(90-p)\$
\$\displaystyle (90p-p^2)\$
\$\displaystyle (90-2p)\$
\$\displaystyle -2p+90=0\$
\$\displaystyle -2p=-90\$
\$\displaystyle p=45\$

I am sure it's something silly considering I am neglecting to incorporate the \$30 into the picture but that's what 6 hours of studying at 3am will do to you! :)
• Apr 20th 2009, 11:35 PM
thelostchild
What you want to do is find the profit, subject to there being a positive demand (price<90) which would be

Profit=Revenue per week - Cost per week

\$\displaystyle
P=pq-30q
\$

\$\displaystyle
P=p(90-p)^2-30(90-p)^2
\$

\$\displaystyle
P=p^3-210p^2+13500p-243000
\$

Find stationary values by taking the first derivative of the profit function and setting it to 0, remembering the above condition then confirm which of the answers is a maximum by finding which would make the second derivative negative
• Apr 20th 2009, 11:52 PM
penbralat
Quote:

Originally Posted by thelostchild
What you want to do is find the profit, subject to there being a positive demand (price<90) which would be

Profit=Revenue per week - Cost per week

\$\displaystyle
P=pq-30q
\$

\$\displaystyle
P=p(90-p)^2-30(90-p)^2
\$

\$\displaystyle
P=p^3-210p^2+13500p-243000
\$

Find stationary values by taking the first derivative of the profit function and setting it to 0, remembering the above condition then confirm which of the answers is a maximum by finding which would make the second derivative negative

Ugh, it's late. Okay, so if I take P'(x) I get \$\displaystyle 3p^2-410p+13500 = 0\$, correct? Forgive me, but refresh me on how exactly to find the specific stationary values. Also, if I take the derivative, again, of the above profit function it would be \$\displaystyle 6p-410=0\$; is that right?
• Apr 21st 2009, 02:14 AM
thelostchild
It must be late, you differentiated it incorrectly it's

\$\displaystyle
P'(p)=3p^2-420p+13500=0\$

\$\displaystyle
p^2-140p+4500=0\$

\$\displaystyle
(p-90)(p-50)=0\$

so the stationary values would be p=90 and p=50

if you plug them into the second differential \$\displaystyle P''(p)=6p-420\$

you find that its negative for p=50 and positive for p=90

so it's maximised for p=50