a company models its revenue as $r(x)=50x - 2x^{2}$ and expenses as $e(x)=10x + 100$. determine the maximum…

a company models its revenue as $r(x)=50x - 2x^{2}$ and expenses as $e(x)=10x + 100$. determine the maximum profit.\na. $x = 10$\nb. $x = 25$\nc. $x = 15$\nd. $x = 20$

a company models its revenue as $r(x)=50x - 2x^{2}$ and expenses as $e(x)=10x + 100$. determine the maximum profit.\na. $x = 10$\nb. $x = 25$\nc. $x = 15$\nd. $x = 20$

Answer

  1. First, find the profit - function (P(x)):
    • The profit function is given by (P(x)=R(x)-E(x)).
    • Given (R(x) = 50x - 2x^{2}) and (E(x)=10x + 100), then (P(x)=(50x - 2x^{2})-(10x + 100)=-2x^{2}+40x - 100).
  2. Then, for a quadratic function (y = ax^{2}+bx + c) (in our case, (a=-2), (b = 40), (c=-100)), the x - value of the vertex (which gives the maximum or minimum of the quadratic function) is (x=-\frac{b}{2a}).
    • Substitute (a=-2) and (b = 40) into the formula (x =-\frac{b}{2a}), we get (x=-\frac{40}{2\times(-2)} = 10).
  3. Now, find the maximum - profit by substituting (x = 10) into the profit function (P(x)):
    • (P(10)=-2\times(10)^{2}+40\times10 - 100=-200 + 400-100 = 100).

The question is asking for the value of (x) that maximizes the profit. The value of (x) for which the profit is maximized is (x = 10).

Answer:

a. (x = 10)