Question Statement

A company determines that for the production of $x$ units of a commodity, its revenue and cost functions are, respectively:

$R(x)=-3x^2+970x$

$G(x)=2x^2+500$

Find:

1. The maximum profit
2. The minimum average cost

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Background and Explanation

This problem involves optimization using calculus, specifically finding extrema (maximums and minimums) of economic functions. You need to construct the profit function from revenue and cost, then use the first and second derivative tests to find optimal production levels.

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Solution

Finding Maximum Profit

First, we define the profit function $P(x)$ as revenue minus cost. Note that we denote the cost function as $C(x)=2x^2+500$ (given as $G(x)$ in the problem):

\begin{align*} P(x) &= R(x) - C(x) \\ &= (-3x^{2} + 970x) - (2x^{2} + 500) \\ P(x) &= -5x^{2} + 970x - 500 \end{align*}

To find the maximum, we take the first derivative with respect to $x$:

$\frac{dP}{dx}=-10x+970$

For critical values, we set the first derivative equal to zero:

\begin{align*} -10x + 970 &= 0 \\ -10x &= -970 \\ x &= 97 \end{align*}

To confirm this is a maximum, we use the second derivative test:

$\frac{d^{2}P}{d{x}^2}=-10$

Since $\frac{d^{2}P}{d{x}^2}=-10<0$, the function is concave down at $x=97$, confirming that $P(x)$ is maximum at $x=97$ units.

Now we calculate the maximum profit by substituting $x=97$ into equation (1):

\begin{align*} P(97) &= -5(97)^{2} + 970(97) - 500 \\ &= -5(9409) + 94090 - 500 \\ &= -47045 + 94090 - 500 \\ P(97) &= 46545 \end{align*}

Maximum Profit = $46,545$

Finding Minimum Average Cost

The average cost function is defined as total cost divided by the number of units:

\begin{align*} C_{\text{avg}}(x) &= \frac{C(x)}{x} \\ &= \frac{2x^{2} + 500}{x} \\ C_{\text{avg}}(x) &= 2x + \frac{500}{x} \end{align*}

Differentiate with respect to $x$:

$\frac{d}{dx}(C_{\text{avg}})=2-\frac{500}{x^2}$

For the second derivative (needed for the optimization test):

\begin{align*} \frac{d^{2}}{dx^{2}}(C_{\text{avg}}) &= \frac{d}{dx}\left(2 - 500x^{-2}\right) \\ &= 0 - 500(-2)x^{-3} \\ \frac{d^{2}}{dx^{2}}(C_{\text{avg}}) &= \frac{1000}{x^{3}} \end{align*}

To find critical values, set the first derivative equal to zero:

\begin{align*} 2 - \frac{500}{x^{2}} &= 0 \\ 2 &= \frac{500}{x^{2}} \\ x^{2} &= \frac{500}{2} = 250 \\ x &= \sqrt{250} = \sqrt{25 \times 10} \\ x &= 5\sqrt{10} \approx 15.81 \end{align*}

Verify this is a minimum using the second derivative test. At $x=5\sqrt{10}\approx 15.81$:

$\frac{d^2}{d{x}^2}(C_{\text{avg}})=\frac{1000}{(15.81{)}^3}>0$

Since the second derivative is positive, $C_{\text{avg}}$ is minimum at $x\approx 15.81$ units.

Calculate the minimum average cost by substituting $x=15.81$ into equation (2):

\begin{align*} C_{\text{avg}}(15.81) &= 2(15.81) + \frac{500}{15.81} \\ &= 31.62 + 31.62 \\ &= 63.24 \end{align*}

Minimum Average Cost = $63.24$ (or exactly $20\sqrt{10}$)

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Key Formulas or Methods Used

• Profit Function: $P(x)=R(x)-C(x)$ (Revenue minus Cost)
• Average Cost: $C_{\text{avg}}(x)=\frac{C(x)}{x}$
• First Derivative Test: Set $\frac{dy}{dx}=0$ to find critical points
• Second Derivative Test:
  • If $\frac{d^{2}y}{d{x}^2}<0$ at critical point $\Rightarrow$ Local Maximum
  • If $\frac{d^{2}y}{d{x}^2}>0$ at critical point $\Rightarrow$ Local Minimum
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

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Summary of Steps

1. Construct Profit Function: Subtract cost from revenue: $P(x)=(-3x^2+970x)-(2x^2+500)=-5x^2+970x-500$
2. Find Critical Point for Profit: Differentiate $P(x)$, set equal to zero: $-10x+970=0\Rightarrow x=97$
3. Confirm Maximum: Check second derivative $P^{\prime \prime }(x)=-10<0$, confirming maximum at $x=97$
4. Calculate Max Profit: Substitute $x=97$ into profit function to get $46,545$
5. Set Up Average Cost: Divide cost by $x$: $C_{\text{avg}}(x)=\frac{2x^2+500}{x}=2x+\frac{500}{x}$
6. Find Critical Point for Avg Cost: Differentiate and set to zero: $2-\frac{500}{x^2}=0\Rightarrow x=\sqrt{250}\approx 15.81$
7. Confirm Minimum: Check second derivative $\frac{1000}{x^3}>0$ at $x=15.81$, confirming minimum
8. Calculate Min Average Cost: Substitute $x=15.81$ to get $63.24$