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The monthly revenue \(R\) (in hundreds of dollars) realized in the sale of Royal electric shavers is related to the unit price \(p\) (in dollars) by the equation $$ R(p)=-\frac{1}{2} p^{2}+30 p $$ a. Sketch the graph of \(R\). b. At what unit price is the monthly revenue maximized?

Short Answer

Expert verified
a. The graph of R(p) is a parabola opening downwards with the vertex at \( (30, 450) \). b. The unit price that maximizes the monthly revenue is 30 dollars.

Step by step solution

01

The function R(p) is a quadratic function given by: \[R(p) = -\frac{1}{2}p^2 + 30p\] #Step 2: Find the vertex of the quadratic function#

The vertex of a quadratic function, \(f(x) = ax^2 + bx + c\), can be found using the formula: \[x = \frac{-b}{2a}\] In our case, \(a = -\frac{1}{2}\) and \(b = 30\). Plug in the values to find the x-coordinate of the vertex, which represents the unit price p: \[p = \frac{-30}{2(-\frac{1}{2})}\] \[p = \frac{-30}{-1}\] \[p = 30\] Now, we can find the corresponding y-coordinate of the vertex, which is the monthly revenue R(p): \[R(30) = -\frac{1}{2}(30)^2 + 30(30)\] \[R(30) = -\frac{1}{2}(900) + 900\] \[R(30) = -450 + 900\] \[R(30) = 450\] So, the vertex of the quadratic function is at point (30, 450). #Step 3: Sketch the graph of R(p)#
02

Based on the vertex (30, 450) and knowing the function is facing downwards (as a < 0), we can sketch a parabolic graph with the peak at the vertex. The axis of symmetry would be at p = 30. #Step 4: Determine the unit price that maximizes the revenue#

Since the vertex represents the maximum point in this case (as the parabola is opening downwards), the unit price that maximizes the revenue is p = 30 dollars. The answer is: a. The graph of R(p) is a parabola opening downwards with the vertex at (30, 450). b. The unit price that maximizes the monthly revenue is 30 dollars.

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Key Concepts

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

Revenue Maximization
In the context of selling products, like the Royal electric shavers from our exercise, *revenue maximization* is an important goal. Here, revenue refers to the total income generated from selling a specific number of products at a certain price. The formula provided for revenue is a quadratic function with respect to unit price \( p \). This can be a powerful tool to determine the best pricing strategy.

With this quadratic revenue function \( R(p) = -\frac{1}{2}p^2 + 30p \), the goal is to find the price \( p \) that generates the highest possible revenue. Since this function involves \( p^2 \), it forms a parabola when graphed, having a peak point called the vertex. In business terms, this peak represents the maximum possible revenue.

To maximize revenue, we look for the vertex of this parabola. Since the coefficient in front of \( p^2 \) is negative, this parabola opens downwards and the vertex is indeed the maximum point of the graph. This emphasizes why understanding the structure of quadratic functions is crucial when making decisions about pricing to achieve the highest revenue possible.
Vertex of a Parabola
The *vertex* of a parabola is a central concept in understanding quadratic functions. It represents the highest or lowest point on the graph, depending on the direction the parabola opens. For a quadratic equation in the form \( f(x) = ax^2 + bx + c \), the vertex \( (h, k) \) can be found using the formula \( h = \frac{-b}{2a} \).

In our example, we have \( a = -\frac{1}{2} \) and \( b = 30 \). Plugging these values into the formula gives \( p = \frac{-30}{2(-\frac{1}{2})} = 30 \). This tells us the unit price that maximizes revenue is 30 dollars. To find the revenue at this price, we substitute \( p = 30 \) back into the original equation: \( R(30) = -\frac{1}{2}(30)^2 + 30(30) = 450 \), giving us a revenue of 450 (in hundreds of dollars).

Knowing how to find the vertex is essential for problems of this nature, as it enables us to solve maximization and minimization problems efficiently. The vertex provides direct insight into the optimal solutions related to many real-life contexts, such as profit maximization, minimum costs, and other economic calculations.
Graph Sketching
*Sketching the graph* of a quadratic function involves understanding its key features, which can include the axis of symmetry, the vertex, and the direction it opens. In our case, the function \( R(p) = -\frac{1}{2}p^2 + 30p \) is a quadratic function that creates a parabolic shape.
  • **Direction**: Since the coefficient of \( p^2 \) is negative, the parabola opens downwards.
  • **Vertex**: We've already determined it is at \( (30, 450) \), making this the highest point on the graph.
  • **Axis of Symmetry**: This is a vertical line that passes through the vertex; here, it is \( p = 30 \).

When sketching, start by marking the vertex on the graph paper. Then draw the axis of symmetry as a dashed line. Since the parabola opens downwards, sketch symmetrical arcs going away from the vertex to represent the downward curve. Understanding these elements will help accurately sketch the parabola, allowing us to visualize the function's behavior over different price points. Such visual aids can be incredibly useful in both academic and practical applications of quadratic functions.

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