# Thread: Elasticity question: Finding maximized price.

1. ## 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 $30 into the picture but that's what 6 hours of studying at 3am will do to you! 2. What you want to do is find the profit, subject to there being a positive demand (price<90) which would be Profit=Revenue per week - Cost per week$\displaystyle
P=pq-30q
\displaystyle
P=p(90-p)^2-30(90-p)^2
\displaystyle
P=p^3-210p^2+13500p-243000
$Find stationary values by taking the first derivative of the profit function and setting it to 0, remembering the above condition then confirm which of the answers is a maximum by finding which would make the second derivative negative 3. Originally Posted by thelostchild What you want to do is find the profit, subject to there being a positive demand (price<90) which would be Profit=Revenue per week - Cost per week$\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$; is that right? 4. It must be late, you differentiated it incorrectly it's$\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