Problem 66 - Algebra and Trigonometry

Question

Sales 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: Identify Profit Equation

The profit equation given in the problem is \[ P(x) = -0.001x^2 + 3x - 1800 \] This represents a quadratic equation in standard form \( ax^2 + bx + c \), where \( a = -0.001 \), \( b = 3 \), and \( c = -1800 \).

Step 2: Determine the Vertex Formula

For a quadratic equation of the form \( ax^2 + bx + c \), the x-coordinate of the vertex, which gives us the point of maximum or minimum value for the parabola, is provided by the formula:\[ x = \frac{-b}{2a} \]

Step 3: Calculate the X-Value for Maximum Profit

Substitute the values of \( a \) and \( b \) into the vertex formula:\[ x = \frac{-3}{2(-0.001)} = \frac{-3}{-0.002} = 1500 \]This means the vendor should sell \( 1500 \) cans to achieve the maximum profit.

Step 4: Calculate Maximum Profit

Substitute \( x = 1500 \) back into the profit equation to find the maximum profit:\[ P(1500) = -0.001(1500)^2 + 3(1500) - 1800 \]First, calculate \( 1500^2 = 2250000 \), then multiply:\[ -0.001 \times 2250000 = -2250 \]Next, calculate:\[ 3 \times 1500 = 4500 \]Now substitute back to find \( P(1500) \):\[ P(1500) = -2250 + 4500 - 1800 \]Simplify to find the maximum profit:\[ P(1500) = 450 \]

Step 5: Conclusion

The maximum profit that the vendor can achieve per day is $450, by selling 1500 cans of soda.

Key Concepts

Maximum Profit

To understand the concept of maximum profit, imagine running a business where you sell items daily. Your profit depends not only on how much you sell but also on factors like cost and demand. The goal is to find the maximum profit you can earn by adjusting the quantity sold.
In mathematical terms, the maximum profit is the highest point on a curve that represents your profit equation. For the vendor in our exercise, the profit equation is a quadratic one:

• \[ P(x) = -0.001x^2 + 3x - 1800 \]

This equation is in the form

• \( ax^2 + bx + c \)

The graph of that equation is a parabola opening downwards, indicating that there will be a single highest point. That point represents the maximum profit. By finding the maximum profit, we understand the optimal conditions and quantity needed to earn the highest possible returns.

Vertex of a Parabola

In the context of quadratic equations, the vertex is a key point that helps determine the maximum or minimum value of the equation. For the vendor's profit equation, the vertex represents the maximum profit point since the parabola opens downwards.
The vertex formula provides the x-coordinate of this critical point, given by:

• \( x = \frac{-b}{2a} \)

Substituting the coefficients from the profit equation:

• \[ x = \frac{-3}{2(-0.001)} = 1500 \]

This x-value tells us how many cans the vendor must sell to achieve maximum profit. Calculating the vertex shows us precisely where the top of the curve lies, giving a clear picture of how to act for optimal outcomes.

Profit Optimization

Profit optimization is all about making sure a business operates in a way that maximizes its earnings. In our scenario with the vendor, profit optimization involves tweaking the number of soda cans sold to hit the maximum profit.
The solution to the exercise shows that the vendor reaches the highest profit when 1500 cans are sold, and the profit maxes out at $450. Using the calculated vertex, businesses can ensure they aren't just aiming for high sales but also the most effective sales amounts that will lead to greater income without unnecessary expenses.
Achieving profit optimization means maintaining a fine balance between

• production costs,
• nick demand,
• and pricing strategies,

which are all reflected in the structure of the quadratic equation. This guide can serve as a crucial decision-making tool for businesses looking to improve profitability through calculated and strategic measures.