Problem 67 - Calculus Early Transcendentals: Pearson New International Edition

Question

The ZEE Company makes zingos, which it markets at a price of \(p(x)=10-0.001 x\) dollars, where \(x\) is the number produced each month. Its total monthly cost is \(C(x)=200+4 x-0.01 x^{2}\). At peak production, it can make 300 units. What is its maximum monthly profit and what level of production gives this profit?

Step-by-Step Solution

Answer

Maximum profit is $618.18 at 273 units produced.

Step 1: Understand Profit Function

First, understand that the company's profit, \( P(x) \), is the difference between total revenue and total cost. The revenue function is calculated by multiplying the price function \( p(x) \) by the number of units \( x \):\[R(x) = x \cdot p(x) = x(10 - 0.001x) = 10x - 0.001x^2\]The profit function is:\[P(x) = R(x) - C(x) = (10x - 0.001x^2) - (200 + 4x - 0.01x^2)\]

Step 2: Simplify the Profit Function

Now simplify the expression for \( P(x) \):\[P(x) = 10x - 0.001x^2 - 200 - 4x + 0.01x^2 = 6x - 0.011x^2 - 200\]This is the profit function that we need to maximize.

Step 3: Find Critical Points

To find the maximum profit, take the derivative of \( P(x) \) and set it to zero:\[P'(x) = 6 - 0.022x\]Set this to zero and solve for \( x \):\[6 - 0.022x = 0 \0.022x = 6 \x = \frac{6}{0.022} \approx 272.73\]Since \( x \) represents the number of units, it should be an integer, so check at \( x = 272 \) and \( x = 273 \).

Step 4: Verify the Second Derivative Test for Maximum

Compute the second derivative \( P''(x) \) to confirm that \( x = 273 \) is a maximum:\[P''(x) = -0.022\]Since \( P''(x) < 0 \), this indicates a concave down curve, confirming that the critical point is a maximum.

Step 5: Compute Maximum Profit

Plug \( x = 273 \) into the profit function \( P(x) \) to find maximum profit:\[P(273) = 6(273) - 0.011(273)^2 - 200\]Calculate:\[P(273) = 1638 - 0.011(74529) - 200\]\[P(273) = 1638 - 819.819 - 200 = 618.181\]Thus, the maximum profit is approximately $618.18.

Key Concepts

Profit Function

The profit function is a mathematical way to determine how much profit a company makes based on its revenue and costs. In simple terms, profit equals revenue minus costs. For the ZEE Company, to derive the profit function, we need two key components, the revenue function and the cost function.

• Revenue is how much money the company makes from selling products.
• Total cost is how much it spends producing those products.

Given that the revenue function is calculated by multiplying the number of units produced, \(x\), by the price function, \(p(x) = 10 - 0.001x\), we find:\[R(x) = x(10 - 0.001x) = 10x - 0.001x^2\]The cost function, \(C(x)\), is provided as:\[C(x) = 200 + 4x - 0.01x^2\]Finally, by subtracting the cost function from the revenue function, we find the profit function:\[P(x) = R(x) - C(x) = 6x - 0.011x^2 - 200\]

Revenue Function

The revenue function helps us understand the total income from product sales. It is determined by the price per unit times the number of units sold. For ZEE Company, the price function is given by \(p(x) = 10 - 0.001x\), indicating that the price decreases slightly with each additional unit sold.

Thus, the revenue function is expressed as:\[R(x) = x imes p(x) = x(10 - 0.001x)\]Simplifying gives us:\[R(x) = 10x - 0.001x^2\]This quadratic function shows how revenue changes with the increase in production. The coefficient \(-0.001\) indicates the revenue decreases as unit production increases beyond a certain point, due to the decreasing price. This information will be vital in calculating where the maximum profit occurs, as revenue directly impacts profit when costs are factored in.

Critical Points

In calculus, critical points help identify where a function reaches its maximum or minimum. For the ZEE Company's profit function, we discover these by finding where the derivative of the profit function equals zero. This tells us points where the slope of the tangent to the profit curve is zero, i.e., where the function levels out and could switch from increasing to decreasing or vice versa.

The derivative of the profit function, \(P(x) = 6x - 0.011x^2 - 200\), is:\[P'(x) = 6 - 0.022x\]Setting \(P'(x) = 0\) to find critical points:\[6 - 0.022x = 0\]Solving gives:\[x = \frac{6}{0.022} \approx 272.73\]Since the number of units, \(x\), must be whole, we check at \(x = 272\) and \(x = 273\). Critical points pinpoint potential locations for maximum or minimum profit, which we verify using further testing.

Second Derivative Test

The second derivative test helps us confirm whether a critical point is a maximum or a minimum. For ZEE Company, the profit function's second derivative indicates the curvature of the function at the critical points. If the second derivative is negative, it suggests a maximum point.

Given the profit derivative \(P'(x) = 6 - 0.022x\), the second derivative is:\[P''(x) = -0.022\]Since \(P''(x) < 0\), it confirms that the graph of the profit function is concave down at the point, meaning we have a local maximum at \(x = 273\). This test conclusively determines that producing 273 units will yield the maximum profit for ZEE Company, allowing us to calculate the highest profit possible from production.