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Chapter 3: Problem 41

Maximum profit: An automobile manufacturer can produce up to 300 cars per day. The profit made from the sale of these vehicles can be modeled by the function \(P(x)=-10 x^{2}+3500 x-66,000\) where \(P(x)\) is the profit in dollars and \(x\) is the number of automobiles made and sold. Based on this model: a. Find the \(y\) -intercept and explain what it means in this context. b. Find the \(x\) -intercepts and explain what they mean in this context. c. How many cars should be made and sold to maximize profit? d. What is the maximum profit?

Short Answer

Expert verified
a. y-intercept is -66,000. b. x-intercepts are 20, 330. c. 175 cars should be made. d. Maximum profit is $240,250.

Step by step solution

01

Determine the y-intercept

To find the y-intercept of the function \( P(x) = -10x^2 + 3500x - 66,000 \), we set \( x = 0 \). This gives us the profit when no cars are sold. \[ P(0) = -10(0)^2 + 3500(0) - 66,000 = -66,000 \] Therefore, the y-intercept is -66,000, which means if no cars are made and sold, the company would incur a loss of $66,000.
02

Determine the x-intercepts

To find the x-intercepts, set \( P(x) = 0 \) and solve for \( x \). This means finding when the profit is zero: \[ -10x^2 + 3500x - 66,000 = 0 \] We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -10 \), \( b = 3500 \), and \( c = -66,000 \). Calculate: \[ x = \frac{-3500 \pm \sqrt{3500^2 - 4(-10)(-66,000)}}{2(-10)} \] Simplifying inside the square root: \( 3500^2 = 12,250,000 \), and \( 4 \times 10 \times 66,000 = 2,640,000 \), so: \[ x = \frac{-3500 \pm \sqrt{12,250,000 - 2,640,000}}{-20} \] \[ x = \frac{-3500 \pm \sqrt{9,610,000}}{-20} \] \( \sqrt{9,610,000} \approx 3,100 \), so: \[ x = \frac{-3500 \pm 3100}{-20} \] This gives \( x = 20 \) and \( x = 330 \). Therefore, the x-intercepts are \( x = 20 \) and \( x = 330 \), meaning the profit is zero when 20 or 330 cars are made and sold.
03

Calculate the vertex (Maximum Profit)

The vertex of the parabola given by the quadratic function \( P(x) = -10x^2 + 3500x - 66,000 \) represents the maximum profit. The x-value of the vertex can be found using \( x = \frac{-b}{2a} \), where \( a = -10 \) and \( b = 3500 \). So \( x = \frac{-3500}{2(-10)} = 175 \). This means to achieve maximum profit, 175 cars should be made and sold.
04

Calculate the maximum profit

To find the maximum profit, substitute \( x = 175 \) back into the profit function \( P(x) \). Calculate: \[ P(175) = -10(175)^2 + 3500(175) - 66,000 \] Which simplifies to: \[ P(175) = -10(30,625) + 612,500 - 66,000 \] \[ P(175) = -306,250 + 612,500 - 66,000 \] \[ P(175) = 240,250 \] The maximum profit is therefore \$240,250.

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Key Concepts

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

Profit Maximization
Profit maximization in the context of quadratic functions is about finding the maximum point of a parabola that opens downward. In our case, we are analyzing the profit made by an automobile manufacturer which is determined by a quadratic profit function, \[ P(x) = -10x^2 + 3500x - 66,000, \]where \( P(x) \) represents the total profit and \( x \) is the number of cars produced and sold.
For maximization, we're interested in finding the vertex of this parabola since it indicates the maximum profit point. The formula\[ x = \frac{-b}{2a} \]is used to calculate this vertex. By substituting \( a = -10 \) and \( b = 3500 \), we find that maximum profit occurs when 175 cars are made and sold.
This tells us that to achieve the highest profit, the manufacturer should focus production around that quantity. This principle of adjusting production to optimize profit is key in economics and business strategy.
X-Intercepts
The x-intercepts of a quadratic function represent the points where the graph crosses the x-axis. They show us the number of cars produced where the profit would be zero. These intercepts are found by solving the equation \[ P(x) = 0 \] for \( x \).
By using the quadratic formula\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]with \( a = -10 \), \( b = 3500 \), and \( c = -66,000 \), we determine that the x-intercepts are 20 and 330.
This means that if only 20 cars or as many as 330 cars are produced, the profits net to zero. In business terms, this is crucial because it informs the manufacturer of inefficient production levels that might lead to losses.
Vertex of a Parabola
The vertex of a parabola, when dealing with a quadratic function opening downwards, marks either a maximum or minimum point. In our exercise, the vertex indicates the maximum profit point. By employing the formula for the x-value of the vertex: \[ x = \frac{-b}{2a}, \]we calculate that the vertex occurs at \( x = 175 \).
This means producing and selling 175 cars maximizes profit. The y-coordinate of the vertex provides the maximum profit value when substituted back into the function: \[ P(175) = 240,250. \]
Understanding the concept of a parabola's vertex helps in optimizing outcomes, such as profit, in various applications dealing with quadratic relationships.
Y-Intercept
The y-intercept is where the graph of a function crosses the y-axis. In quadratic functions, it shows the output value when the input \( x \) is zero. For the profit function \[ P(x) = -10x^2 + 3500x - 66,000, \]the y-intercept is found by evaluating \[ P(0). \]This calculation simplifies to \( P(0) = -66,000 \).
This tells us that if no cars are produced and sold, the company incurs a loss of $66,000, likely due to fixed costs.
  • Understanding the y-intercept is valuable for business, as it highlights potential losses at zero production.
By managing production levels, a company can aim to move away from this loss scenario towards profit.

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Most popular questions from this chapter

A large group of students is asked to memorize a list of 50 Italian words, a language that is unfamiliar to them. The group is then tested regularly to see how many of the words are retained over a period of time. The average number of words retained is modeled by the function \(W(t)=\frac{6 t+40}{t},\) where \(W(t)\) represents the number of words remembered after \(t\) days. a. Graph the function over the interval \(t \in[0,40] .\) How many days until only half the words are remembered? How many days until only one-fifth of the words are remembered? b. After 10 days, what is the average number of words retained? How many days until only 8 words can be recalled? c. What is the significance of the horizontal asymptote (what does it mean in this context)?

The surface area of a spherical cap is given by \(S=2 \pi r h,\) where \(r\) is the radius of the sphere and \(h\) is the perpendicular distance from the sphere's surface to the plane intersecting the sphere, forming the cap. The volume of the cap is \(V=\frac{1}{3} \pi h^{2}(3 r-h) .\) Similar to Exercise \(61,\) a formula can be found that will minimize the area of a cap that holds a specified volume. a. Solve the volume formula for the variable \(r\) b. Substitute the resulting expression for \(r\) into the surface area formula and simplify. The result is a formula for surface area given solely in terms of the volume \(V\) and the height \(h\). c. Assume the volume of the spherical cap is \(500 \mathrm{cm}^{3} .\) Use a graphing calculator to graph the resulting function on an appropriate window, and use the graph to find the height \(h\) that will result in a spherical cap with the smallest possible area, while still holding a volume of \(500 \mathrm{cm}^{3}\) d. Use this value of \(h\) and \(V=500 \mathrm{cm}^{3}\) to find the radius of the sphere.

Solve each quadratic inequality by locating the \(x\) -intercept(s) (if they exist), and noting the end behavior of the graph. Begin by writing the inequality in function form as needed. $$-x^{2}>2$$

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