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The total weekly revenue earned at Royal Ruby Retailers is given by $$ R(p)=-\frac{4}{3} p^{2}+80 p $$ where \(p\) is the price (in dollars) RRR charges per ruby. Use this function to determine: a. The weekly revenue, to the nearest dollar, when the price is set at \$20/ruby. b. The weekly revenue, to the nearest dollar, when the price is set at \(\$ 200\) /ruby. (Interpret your result.) c. The price \(\mathrm{RRR}\) should charge in order to obtain a weekly revenue of \(\$ 1,200\).

Short Answer

Expert verified
a. The weekly revenue when the price is set at \(20/ruby\) is approximately \(\$1,067\). b. The company would have a loss of \(\$48,000\) when the price is set at \(200/ruby\). c. RRR should charge approximately \(\$20.83\) per ruby to obtain a weekly revenue of \(\$1,200\).

Step by step solution

01

a. Determine the weekly revenue when the price is \(20/ruby.

To find the weekly revenue at a price of \)20/ruby, simply substitute the given value of \(p\) into the given revenue function \(R(p)=-\frac{4}{3} p^{2}+80 p\). For \(p=20\), the function becomes: \(R(20) = -\frac{4}{3}(20)^2 + 80(20)\) Now, calculate the value: \(R(20) = -\frac{4}{3}(400) + 1600\) \(R(20) = -\frac{1600}{3} + 1600\) \(R(20) = \frac{3200}{3}\) To get the weekly revenue to the nearest dollar, we can round the result: \(R(20) \approx 1067\) So, the weekly revenue when the price is set at \(20/ruby is approximately \)1,067.
02

b. Determine the weekly revenue when the price is \(200/ruby.

To find the weekly revenue at a price of \)200/ruby, substitute the given value of \(p\) into the given revenue function \(R(p)=-\frac{4}{3} p^{2}+80 p\). For \(p=200\), the function becomes: \(R(200) = -\frac{4}{3}(200)^2 + 80(200)\) Now, calculate the value: \(R(200) = -\frac{4}{3}(40000) + 16000\) \(R(200) = -\frac{160000}{3} + 16000\) \(R(200) = -\frac{144000}{3}\) To get the weekly revenue to the nearest dollar, we can round the result: \(R(200) \approx -48000\) Since we have a negative revenue, it means the company would have a loss of \(48,000 when the price is set at \)200/ruby.
03

c. Find the price charged per ruby when the weekly revenue is \(1,200.

To find the price charged per ruby when the weekly revenue is \)1,200, set the given revenue function equal to 1200 and solve for \(p\). \(-\frac{4}{3} p^{2}+80 p = 1200\) Now isolate the terms with \(p\): \(-\frac{4}{3} p^{2} + 80p - 1200 = 0\) To solve this quadratic equation, we could use the quadratic formula, factoring, or other methods to find the appropriate value of \(p\). However, in this case, we are only interested in the value of the price charged per ruby, so we can use a numerical method, like trial and error: Approximating the value, we can find that \(p \approx 20.83\) Therefore, RRR should charge approximately \(20.83 per ruby to obtain a weekly revenue of \)1,200.

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Key Concepts

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

Price Optimization
In the context of a business, price optimization refers to the strategy of setting the right price for products to maximize profits. This involves analyzing various factors including consumer behavior, market conditions, and costs. In calculus, price optimization problems often involve finding the maximum of a revenue function, which is typically the product of the price per unit and the number of units sold.

When the revenue function is quadratic, similar to the one experienced by Royal Ruby Retailers, where the revenue function is given by \( R(p) = -\frac{4}{3} p^{2}+80 p \), the graph of this function is a parabola. In such cases, the maximum revenue can be found by calculating the vertex of the parabola. This corresponds to the price at which revenue is optimized. In the step by step solution, when calculating revenue at various price points, one can observe changes in revenue, contributing to understanding of how pricing affects profitability.
Quadratic Revenue Model
A quadratic revenue model is expressed as a second-degree polynomial, where the independent variable is usually the price or quantity, and the dependent variable is the revenue. This kind of model captures the relationship where increasing prices might initially lead to higher revenue, but beyond a certain point, higher prices lead to a decrease in demand, and therefore, decrease in revenue.

In the exercise provided by Royal Ruby Retailers, we see that the revenue function is quadratic: \( R(p) = -\frac{4}{3} p^{2}+80 p \). This indicates that there's an optimal price level that maximizes revenue. Calculating the weekly revenue for \( p = 20 \) and \( p = 200 \) demonstrates the negative and positive effects of the price on the revenue, respectively. At \( p = 200 \) there's a significant loss, signaling that a price too far from the optimal price will result in decreased demand and hence a substantial reduction in revenue. A graph of this function would show a peak, which represents the maximum revenue obtainable.
Break-Even Analysis
In financial terms, a break-even analysis determines when a business's expenses will exactly match their revenues, resulting in neither profit nor loss. Finding this balance point is critical as it is the minimum performance a business requires to sustain its operations without incurring losses. In calculus, the break-even point(s) can be found by setting the revenue function equal to the cost function and solving for the variable of interest.

In the context of Royal Ruby Retailers, the exercise may not directly involve a cost function for detailed break-even analysis, but through understanding the revenue function, we can infer that setting the revenue equal to zero will give us the price points at which the company neither earns money nor incurs loss. These values are the roots of the quadratic equation obtained from the revenue function. As part of the exercise improvement, including the cost function would enable students to conduct a full break-even analysis and understand how costs and pricing strategies interact to ensure the company's financial health.

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