Find the value of x that maximizes y = 20x – x Find the value of x that minimizes y = x Given TC = x Given TR = 100x – x A firm's profit function is π = 120x – x

Here are the solutions to the practice problems.

41. Find the value of $x$ that maximizes $y = 20x - x^2$.

Step 1: To find the maximum value, we take the first derivative of the function $y$ with respect to $x$. $\frac{dy}{dx} = \frac{d}{dx}(20x - x^2) = 20 - 2x$

Step 2: Set the first derivative equal to zero and solve for $x$. $20 - 2x = 0$ $2x = 20$ $x = 10$ The value of $x$ that maximizes $y$ is $\boxed{10}$.

42. Find the value of $x$ that minimizes $y = x^2 + 4x + 7$.

Step 1: To find the minimum value, we take the first derivative of the function $y$ with respect to $x$. $\frac{dy}{dx} = \frac{d}{dx}(x^2 + 4x + 7) = 2x + 4$

Step 2: Set the first derivative equal to zero and solve for $x$. $2x + 4 = 0$ $2x = -4$ $x = -2$ The value of $x$ that minimizes $y$ is $\boxed{-2}$.

43. Given $TC = x^2 + 10x + 50$, find the Marginal Cost.

Step 1: Marginal Cost (MC) is the derivative of the Total Cost (TC) function with respect to $x$. $MC = \frac{d(TC)}{dx}$

Step 2: Differentiate the given TC function. $MC = \frac{d}{dx}(x^2 + 10x + 50) = 2x + 10$ The Marginal Cost is $\boxed{2x + 10}$.

44. Given $TR = 100x - x^2$, find the Marginal Revenue.

Step 1: Marginal Revenue (MR) is the derivative of the Total Revenue (TR) function with respect to $x$. $MR = \frac{d(TR)}{dx}$

Step 2: Differentiate the given TR function. $MR = \frac{d}{dx}(100x - x^2) = 100 - 2x$ The Marginal Revenue is $\boxed{100 - 2x}$.

45. A firm's profit function is $\pi = 120x - x^2 - 200$. Determine the output level that maximizes profit.

Step 1: To maximize profit, we take the first derivative of the profit function $\pi$ with respect to $x$. $\frac{d\pi}{dx} = \frac{d}{dx}(120x - x^2 - 200) = 120 - 2x$

Step 2: Set the first derivative equal to zero and solve for $x$. $120 - 2x = 0$ $2x = 120$ $x = 60$ The output level that maximizes profit is $\boxed{60}$.

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