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Chapter 2: Problem 40

The graph of each function has one relative extreme point. Find it (giving both \(x\) - and \(y\) -coordinates) and determine if it is a relative maximum or a relative minimum point. Do not include a sketch of the graph of the function. $$f(x)=30 x^{2}-1800 x+29,000$$

Short Answer

Expert verified
The relative minimum point is at \((30, 2000)\).

Step by step solution

01

- Find the First Derivative

To find the extreme point, start by calculating the first derivative of the function. The function given is: \[ f(x) = 30x^2 - 1800x + 29000 \] The first derivative is: \[ f'(x) = \frac{d}{dx} (30x^2 - 1800x + 29000) \] Simplify the derivative: \[ f'(x) = 60x - 1800 \]
02

- Set the First Derivative to Zero

Set the first derivative equal to zero to find the critical point(s): \[ 60x - 1800 = 0 \] Solve for \(x\): \[ 60x = 1800 \] \[ x = 30 \]
03

- Find the Second Derivative

Calculate the second derivative of the function to determine if the critical point is a maximum or minimum: \[ f''(x) = \frac{d}{dx} (60x - 1800) \] Simplify the second derivative: \[ f''(x) = 60 \]
04

- Determine the Nature of the Extreme Point

Use the second derivative test: if \(f''(x) > 0\), the critical point is a relative minimum; if \(f''(x) < 0\), it is a relative maximum. Since \( f''(30) = 60 > 0 \), the critical point at \(x = 30\) is a relative minimum.
05

- Find the y-coordinate of the Extreme Point

Substitute \(x = 30\) back into the original function to find the \(y\)-coordinate: \[ f(30) = 30(30)^2 - 1800(30) + 29000 \] \[ f(30) = 30(900) - 54000 + 29000 \] \[ f(30) = 27000 - 54000 + 29000 \] \[ f(30) = 2000 \]
06

- State the Relative Extreme Point

From the calculations, the relative minimum point is at \((30, 2000)\).

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Key Concepts

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

First Derivative
The first derivative of a function is a crucial tool in calculus. It gives us the rate at which a function's value is changing at any point. To find the first derivative of the function given, we need to differentiate the function with respect to \(x\). Here's the original function:

\[ f(x) = 30x^2 - 1800x + 29000 \]

When we differentiate it, using standard differentiation rules, we get:

\[ f'(x) = 60x - 1800 \]

Setting the first derivative to zero helps us find the critical points, which are potential points for relative maxima or minima. So, we solve:

\[ 60x - 1800 = 0 \]

and

\[ x = 30 \]

This \(x\)-value represents a critical point. But to understand the nature of this point, whether it is a maximum or a minimum, we need more information.
Second Derivative
The second derivative of a function gives us information about the curvature of the function. To find the second derivative, we differentiate the first derivative:

\[ f''(x) = \frac{d}{dx} (60x - 1800) = 60 \]

The second derivative is constant, which simplifies the analysis. The second derivative test states that if the second derivative at a critical point is positive, the function has a relative minimum at that point. If it is negative, the function has a relative maximum. Here, we have:

\[ f''(30) = 60 > 0 \]

Since the second derivative is positive, the function \( f(x) = 30x^2 - 1800x + 29000 \) has a relative minimum at \( x = 30 \).
Relative Minimum
A relative minimum is a point where the function's value is lower than all other nearby values. From the earlier steps, we determined that the function has a relative minimum at \( x = 30 \). To find the \( y \)-coordinate of this minimum, we substitute \( x = 30 \) back into the original function:

\[ f(30) = 30(30)^2 - 1800(30) + 29000 \]

Simplifying this,

\[ f(30) = 30 \times 900 - 54000 + 29000 \] \[ f(30) = 27000 - 54000 + 29000 \] \[ f(30) = 2000 \]

So, the coordinates of the relative minimum point are \((30, 2000)\). This is the point where the function achieves its lowest value in the neighborhood of \( x = 30 \).

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Most popular questions from this chapter

Let \(Q(x)\) denote the total cost of manufacturing \(x\) units of some product. Then \(C(x)\) is an increasing function for all \(x\). For small values of \(x\), the rate of increase of \(C(x)\) decreases (because of the savings that are possible with "mass production"). Eventually, however, for large values of \(x\), the cost \(Q(x)\) increases at an increasing rate. (This happens when production facilities are strained and become less efficient.) Sketch a graph that could represent \(Q(x).\)

Sketch the following curves, indicating all relative extreme points and inflection points. $$y=2 x^{3}-3 x^{2}-36 x+20$$

The demand equation for a company is \(p=200-3 x,\) and the cost function is $$Q(x)=75+80 x-x^{2}, \quad 0 \leq x \leq 40.$$ (a) Determine the value of \(x\) and the corresponding price that maximize the profit. (b) If the government imposes a tax on the company of \(\$ 4\) per unit quantity produced, determine the new price that maximizes the profit. (c) The government imposes a tax of \(T\) dollars per unit quantity produced (where \(0 \leq T \leq 120\) ), so the new cost function is $$C(x)=75+(80+T) x-x^{2}, \quad 0 \leq x \leq 40.$$ Determine the new value of \(x\) that maximizes the company's profit as a function of \(T .\) Assuming that the company cuts back production to this level, express the tax revenues received by the government as a function of \(T\). Finally, determine the value of \(T\) that will maximize the tax revenue received by the government.

Draw the graph of a function \(y=f(x)\) with the stated properties. The function decreases and the slope increases as \(x\) increases. [Note: The slope is negative but becomes less negative.]

The monthly demand equation for an electric utility company is estimated to be $$p=60-\left(10^{-5}\right) x,$$ where \(p\) is measured in dollars and \(x\) is measured in thousands of kilowatt- hours. The utility has fixed costs of 7 million dollars per month and variable costs of \(\$ 30\) per 1000 kilowatt-hours of electricity generated, so the cost function is $$a(x)=7 \cdot 10^{6}+30 x,$$ (a) Find the value of \(x\) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. (b) Suppose that rising fuel costs increase the utility's variable costs from \(\$ 30\) to \(\$ 40\), so its new cost function is $$C_{1}(x)=7 \cdot 10^{6}+40 x.$$ Should the utility pass all this increase of \(\$ 10\) per thousand kilowatt- hours on to consumers? Explain your answer.
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