A producer of computer graphics software finds that selling price p of its software is related to the number of x copies of its software sold annually by the demand equation, x = 10000 -200p , while its total cost in producing and marketing these x copies is given by the function $$C(x)=50000+5x $$ Find the price p for which profits will be a maximum. Find the maximum profit earned by selling at this price.
My solution is: p(x) = xp(x)-C(x) = x(10000-200p)-(50000-5x) = -200px+9995x-50000
p'(x) = -200p + 9995
let p'(x) = 0
-200p + 9995 = 0
p = -9995/(-200) = 49.975
The answer is supposed to be p = $27.50 and P(x) = $51,250
Please help me I'm stuck at this question
Let $x=$ number of copies, $p=$ price, and $P=$ profit.
Then $x=10000-200p$, so $p(x)=\dfrac{10000-x}{200}=50-\dfrac x{200}$.
$P(x)=x p(x) - C(x)=x\left(50-\dfrac x{200}\right)-(50000+5x)=-\dfrac1{200}x^2+45x-50000.$
$P'(x)=-\dfrac1{100}x+45.$
Can you take it from here?
[Solve for $x$ such that $P'(x)=0$; then evaluate $p(x)$ and $P(x)$.]