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Retail Store Profit The manager of a retail store has figured that her monthly profit \(P\) (in thousands of dollars) is determined by her monthly advertising expense \(x\) (in tens of thousands of dollars) according to the formula $$ P=x^{3}-20 x^{2}+100 x \text { for } 0 \leq x \leq 4 $$ For what value of \(x\) does she get \(\$ 147,000\) in profit?

Short Answer

Expert verified
The value of \(x\) is \(3\).

Step by step solution

01

Understand the Problem

We need to find the value of the advertising expense, represented by \(x\), that will result in a profit \(P\) of \(147\) (since the profit \(P\) is given in thousands of dollars).
02

Set Up the Equation

The profit formula is given by \[ P = x^{3} - 20x^{2} + 100x \]. We need to find \(x\) such that \[ x^{3} - 20x^{2} + 100x = 147 \].
03

Substitute and Simplify

Substitute \(P = 147\) into the profit equation: \[ x^{3} - 20x^{2} + 100x - 147 = 0 \]. This is a cubic equation in terms of \(x\).
04

Solve the Cubic Equation

Use numerical methods or a graphing calculator to find the roots of \[ x^{3} - 20x^{2} + 100x - 147 = 0 \]. After solving, we find that one of the roots is approximately \(x = 3\).
05

Verify the Solution

Substitute \(x = 3\) back into the profit equation to verify: \[ P = 3^{3} - 20(3)^{2} + 100(3) \] \[ P = 27 - 180 + 300 \] \[ P = 147 \] Since the calculated profit matches the given profit, the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Retail Store Profit
The profit of a retail store is the amount of money that remains after all expenses have been subtracted from the total revenue. In our exercise, the store manager has created a formula to determine the monthly profit, given a specific amount spent on advertising. The formula is: \[ P = x^3 - 20x^2 + 100x \] Here, \(P\) represents the profit in thousands of dollars, and \(x\) is the advertising expense in tens of thousands of dollars. By understanding this relationship, the manager can make informed decisions about how much to spend on advertising to achieve desired profit levels.
Advertising Expense
Advertising expense is the amount of money spent on promoting a store's products or services. This can include costs for online ads, print ads, billboards, and more. In our problem, the advertising expense is represented by the variable \(x\). \ Spending on advertising can help increase a store’s revenue by attracting more customers and increasing sales. However, spending too much might not always lead to higher profits if the costs outweigh the benefits.
This is why it is crucial to find the optimal advertising expense that maximizes profit. The optimal value of \(x\) ensures that every dollar spent contributes effectively to the store's profitability.
Profit Maximization
Profit maximization refers to finding the level of output or input that leads to the highest possible profit. In the context of our exercise, it involves determining the best amount of advertising expense (\(x\)) to maximize monthly profit (\(P\)).
To find the optimal value, you may need to solve mathematical equations that represent the relationships between different variables. Here, the profit formula is a cubic equation, \[ x^3 - 20x^2 + 100x - 147 = 0 \] We solve this to find the value of \(x\) that yields a profit (\(P\)) of \$147,000. This process involves substituting \(P = 147\) into the equation and solving for \(x\).
Numerical Methods
Solving cubic equations like ours isn't always straightforward. Analytical solutions can be complex, so we often turn to numerical methods to find an approximate solution. Numerical methods include techniques like Newton's method, bisection method, and using graphing calculators or software.
These methods involve iteratively guessing and checking values of \(x\) until you find one that makes the equation true. For example, in our exercise, we use a numerical method to find the root of the equation \[ x^3 - 20x^2 + 100x - 147 = 0 \] This root, which is approximately \(x = 3\), helps us determine the advertising expense needed to achieve a \$147,000 profit.\ To verify, we substitute \(x = 3\) back into the original profit equation and check if the result matches the expected profit.

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