Question 1

A small business has the following revenue and cost functions for a product: and . If the break-even point lies where revenue equals cost, how many break-even points are there?

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Question 2

A ball is hit upward with an initial velocity of 128 feet per second. After seconds, its height is given by the function . How many seconds does the ball take to reach its maximum height?

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Question 3

A ball is hit upward with an initial velocity of 128 feet per second. After seconds, its height is given by the function . What is the maximum height in feet of the ball?

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Question 4

What value of makes a minimum?

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Question 5

Suppose that a rifle bullet follows a parabolic path given by , where both and are measured in feet. How far downrange will the bullet hit the ground?

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To find the distance downrange, we need to find the horizontal distance the bullet travels before hitting the ground. This occurs when . 
Setting  and solving for , we get:
$$t=\frac{2V_0\sin(	heta)}{g}$$
Substituting this value for  into the equation for  gives:
$$x=\frac{V_0^2}{g}\sin(2	heta)$$
The maximum range (i.e., the distance downrange when the angle of inclination is  ) is obtained when:
$$\sin(2	heta)=1$$
$$\Rightarrow 2	heta=90^\circ$$
$$\Rightarrow 	heta=45^\circ$$
Substituting this value into the above equation gives:
$$x_{max}=\frac{V_0^2}{g}\sin(2	imes45^\circ)$$
We are not given the initial velocity of the bullet, but we can eliminate choices (B), (C), and (D) because they are greater than the maximum range, which occurs at an angle of 45 degrees.

To eliminate choice (A), we need to estimate whether the maximum range of a rifle bullet is likely to exceed 2000 feet. According to several sources, the maximum range of a high-powered rifle bullet is typically around 4000-5000 feet, so it is plausible that the bullet would travel farther than 2000 feet downrange. Therefore, the best choice is (A).

Question 6

What is the maximum value of the function ?

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Question 7

Cutting 7-inch-long squares from a square piece of cardboard and folding up the edges, you want to make an open box (without lid) with a square bottom, a height of 7 inches, and a volume of 252 cubic inches. What should be the dimension of the cardboard?

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Question 8

A walkway of a park has the shape of a right triangle with three side lengths . Find the perimeter of the right triangle.

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Question 9

You have a 400-foot roll of fencing to enclose a playground. If you want to construct a rectangular playground, what is the largest possible area of the playground?

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Question 10

You have a 600-foot roll of fencing to enclose two separate rectangular playgrounds for different age groups. If the two playgrounds are identical and adjacent, and form a larger rectangular region, what is the largest possible area of the playgrounds in total?

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Let the length of each playground be $l$ and the width be $w$. Since the two playgrounds are identical and adjacent, the larger rectangle has length $2l$ and width $w$.

Since there are 600 feet of fencing, we have $2(2l+w)=600$, or $4l+2w=600$. Solving for $w$ in terms of $l$, we get $w=300-2l$, or $l=\frac{1}{2}(300-w)$.

The area of a rectangle is given by $lw$. Substituting for $l$ in terms of $w$, we have that the total area is $2\left(\frac{1}{2}(300-w)\right)w=300w-w^2$.

To find the maximum value of this expression, we complete the square: $$300w-w^2=-\left(w^2-300w \right)=-(w^2-300w+22500-22500)=-(w-150)^2+22500.$$

Since $(w-150)^2 \geq 0$ for all $w$, it follows that $-(w-150)^2 \leq 0$, and hence the maximum value of $300w-w^2$ is $22,\!500$, which occurs when $w=150$.

Plugging this value of $w$ back in, we can solve for the length of each playground: $l=\frac{1}{2}(300-w)=75$. Therefore, the total area of the two playgrounds is $2lw=2(75)(150)=\boxed{22,\!500}$ square feet.

Question 11

A rectangular playground has the dimensions 20 yards by 30 yards. You want to double the space by adding the same amount to both the length and width. What is the perimeter of the expanded playground?

Accepted Answer

Doubling the area of the playground means the new area should be 2 * (20 * 30) = 1200 square yards. Let x be the amount added to both the length and width. The equation for the new area is (20 + x)(30 + x) = 1200. Solving for x gives a quadratic equation, but since we're looking for the perimeter, we don't need to solve for x explicitly. The original perimeter is 2(20 + 30) = 100 yards. Doubling the area approximately doubles the dimensions, so the new dimensions are approximately 40 yards by 60 yards, making the new perimeter 2(40 + 60) = 200 yards. However, since none of the options match 200 yards and considering the need for an exact calculation, let's correct the approach: To double the area without solving for x directly, we notice an error in the initial explanation. Doubling the area doesn't necessarily double the dimensions linearly because we're adding the same amount to both dimensions, not multiplying. The correct approach involves recognizing the mistake in the calculation process that led to the incorrect perimeter estimation. Without the exact value of x, we cannot directly calculate the new perimeter based on the provided options. Therefore, a more detailed calculation or correction is needed to find the right answer. Given the mistake in the initial explanation, the correct answer should be recalculated based on the actual method to double the area and find the new perimeter correctly.

Question 12

For a certain product X, the profit is modeled as , where equals the number of units sold. How many units sold will produce the maximum profit?

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Quiz 3: Linear and Exponential Change: Comparing Growth Rates