Problem 66 - Precalculus Mathematics for Calculus

Question

\(A\) soft-drink vendor at a popular beach analyzes his sales records and finds that if he sells \(x\) cans of soda pop in one day, his profit (in dollars) is given by $$P(x)=-0.001 x^{2}+3 x-1800$$ What is his maximum profit per day, and how many cans must he sell for maximum profit?

Step-by-Step Solution

Answer

The maximum profit is $450, achieved by selling 1500 cans.

Step 1: Understand the Profit Function

The profit function for the vendor is given as \( P(x) = -0.001x^2 + 3x - 1800 \). This is a quadratic function of the form \( ax^2 + bx + c \) where \( a = -0.001 \), \( b = 3 \), and \( c = -1800 \). Since \( a < 0 \), the parabola opens downward, indicating the function has a maximum.

Step 2: Find the Vertex of the Parabola

To find the maximum profit, we need to locate the vertex of the parabola, as it represents the maximum point. The x-value of the vertex can be found using the formula \( x = -\frac{b}{2a} \) from the coefficients of the quadratic equation. Here, \( a = -0.001 \) and \( b = 3 \).

Step 3: Calculate the Optimal Number of Cans

Substitute \( a = -0.001 \) and \( b = 3 \) into the vertex formula to find \( x \): \[ x = -\frac{3}{2(-0.001)} = 1500 \]So, the optimal number of cans to sell for maximum profit is 1500.

Step 4: Determine the Maximum Profit

To find the maximum profit, substitute \( x = 1500 \) back into the profit function \( P(x) \):\[ P(1500) = -0.001(1500)^2 + 3(1500) - 1800 \]Calculate step-by-step:1. \( 1500^2 = 2250000 \)2. \( -0.001 imes 2250000 = -2250 \)3. \( 3 imes 1500 = 4500 \)4. \( -2250 + 4500 - 1800 = 450 \)So, the maximum profit is $450.

Key Concepts

Quadratic Function

Understanding quadratic functions is essential when dealing with profit optimization problems. A quadratic function is typically given in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our case, the quadratic function used in the profit scenario is \( P(x) = -0.001x^2 + 3x - 1800 \). Here, the variable \( x \) represents the quantity of cans sold, while \( P(x) \) indicates the profit in dollars.
Some important characteristics of this function include:

• The coefficient \( a \) is negative (-0.001), causing the parabola to open downwards. This indicates that there is a maximum point in the graph.
• The parabolic shape is symmetric, meaning the line of symmetry runs through the vertex, the highest point on the curve.
• The function represents a real-world scenario where selling too few or too many cans results in less than maximum profit.

Understanding these elements helps you grasp why the quadratic equation is the perfect model for situations when there is an optimal level of activity (here, sales) where profit is maximized.

Vertex of Parabola

The vertex of a parabola is a critical point in the graph of a quadratic function. For the problem at hand, the vertex represents the maximum profit the vendor can achieve by selling a certain number of cans. The vertex provides both the \( x \)-value for optimal sales and the \( y \)-value for the maximum profit.
To find the \( x \)-coordinate of the vertex, use the formula \( x = -\frac{b}{2a} \). In this context:

• The coefficient \( b = 3 \).
• The coefficient \( a = -0.001 \).

Substituting these values into the formula, we find \( x = 1500 \). This means that to achieve the highest profit, the vendor should aim to sell 1500 cans in a day.
The \( x \)-value gives us the key to maximizing sales, but it's the \( y \)-value, \( P(x) \), which gives the profit amount. Vertex analysis helps translate the abstract concepts to actionable decisions for the vendor.

Maximum Profit Calculation

Maximum profit calculation involves substituting the \( x \)-value from the vertex back into the original profit equation to solve for \( P(x) \). With the previously calculated vertex point \( x = 1500 \), we can determine the peak dollar earnings the vendor can achieve in a day.
Here's the calculation process,

• Start with the formula: \( P(x) = -0.001x^2 + 3x - 1800 \).
• Substitute \( x = 1500 \) into \( P(x) \): \( P(1500) = -0.001(1500)^2 + 3(1500) - 1800 \).
• Calculate \( (1500)^2 = 2250000 \).
• Multiply \(-0.001 \times 2250000 = -2250 \).
• Add \( 3 \times 1500 = 4500 \).
• Combine the results: \(-2250 + 4500 - 1800 = 450 \).

Thus, the maximal daily profit the vendor can achieve is $450. This calculation illustrates the importance of understanding and applying the quadratic function and its vertex, providing an essential tool in maximizing profits efficiently.