Question

company uses the function \(P(x)=150 x-x^{2}\) to determine how much profit the company will make when it sells 150 units of a certain product that sells for \(x\) dollars per unit. How much more profit per unit, in dollars, will the company make if it charges \(\$ 25\) for the product than if it charges \(\$ 20 ?\) (Ignore the dollar sign when gridding your response.)

Short answer

The company makes 525 dollars more profit per unit by charging 25 dollars instead of 20 dollars.

Step-by-step solution

Understand the problem

The function given is a profit function, denoted by the equation \(P(x) = 150x - x^2\), which calculates the profit based on the price per unit, \(x\) dollars. The task is to find the difference in profit when the product is sold at \(\$25\) compared to when it is sold at \(\$20\).

Substitute \(x = 25\) into the profit function

Evaluate the profit function at \(x = 25\): \[P(25) = 150(25) - (25)^2\]. Calculate: \[P(25) = 3750 - 625\], so \[P(25) = 3125\].

Substitute \(x = 20\) into the profit function

Evaluate the profit function at \(x = 20\): \[P(20) = 150(20) - (20)^2\]. Calculate: \[P(20) = 3000 - 400\], so \[P(20) = 2600\].

Calculate the difference in profit

Subtract the profit at \(x = 20\) from the profit at \(x = 25\): \[P(25) - P(20) = 3125 - 2600\]. The result is: \[P(25) - P(20) = 525\].

Key concepts

profit function

A profit function is a mathematical equation used to calculate the profit generated by selling a product at a given price. In this exercise, the profit function is denoted as \(P(x) = 150x - x^2\). Here, \(x\) represents the selling price per unit in dollars. The function combines two parts: \(150x\), which indicates the revenue, and \((- x^2)\), which shows the losses or costs associated with producing those units. Using this function helps companies understand how different prices affect their profit.

For instance, if the price is too high or too low, the squared term may significantly affect the profit, leading to lower earnings. This function helps in determining the optimal price point to maximize profit.

substitution method

The substitution method involves replacing variables with specific values to simplify the equation or to find particular outcomes. In this problem, we substitute specific values of \ x \ (unit prices in dollars) into our profit function \(P(x) = 150x - x^2 \).

First, we substitute \ x = 25 \ and then \ x = 20 \ into the function to find out the corresponding profits.

Evaluating the function, \( P(25) = 150(25) - (25)^2 = 3750 - 625 = 3125 \) and \( P(20) = 150(20) - (20)^2 = 3000 - 400 = 2600 \).

By substituting these values, we can easily find the profits for specific unit prices rather than trying to solve the function in general terms.

difference in profit

The difference in profit is simply the amount by which the profit changes when comparing two different scenarios. In this exercise, we're comparing profits at two different prices: \ \(25 \ and \ \)20.

After calculating the profit for both prices, \ P(25) = 3125 \ and \ P(20) = 2600 \, the difference in profit can be found by subtracting the smaller profit from the larger one.

Thus, \( 3125 - 2600 = 525 \). This means the company makes \ \(525 \ more when selling the product at \ \)25 \ per unit compared to \ $20 \ per unit. Understanding the difference in profit helps determine the most lucrative price points for the product.

algebraic calculations

Algebraic calculations are crucial for solving equations and understanding relationships between variables. In this problem, we're using basic algebra to find specific values of the profit function.

Here's a step-by-step breakdown of the calculations:

• First, we substitute \ x = 25 \ into the profit function to get \ P(25) = 150(25) - (25)^2 \.
• Calculate: \ P(25) = 3750 - 625 = 3125 \.
• Then, we substitute \ x = 20 \ into the function to get \ P(20) = 150(20) - (20)^2 \.
• Calculate: \ P(20) = 3000 - 400 = 2600 \.

The final algebraic step is to find the difference:

\(3125 - 2600 = 525 \). This clear sequence of operations demonstrates how algebra facilitates problem solving in practical scenarios, like determining profit differences.