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Practice Problems

Practice Problem #1

The daily profit, P, of a cotton candy vendor at the fair is described by the function P = −60x2+ 240x − 80, where x dollars is the selling price of a bag of cotton candy. Use this equation to answer the questions below.

  1. What is the maximum daily profit?
  2. How much should each bag of cotton candy cost to generate the maximum profit?
  3. How much profit would be generated if the cotton candy sold for $1.50?

1. What is the maximum daily profit?

To find the maximum daily profit, you will want to find the maximum point of the function.

Hint: The maximum point is also the vertex.

x equals negative b over 2 a. = negative blank over two times blank

Simplify

negative two hundred and forty over negative one hundred and twenty=

Substitute x = 2 into the equation and simplify.

P = −60x2+ 240x − 80

P = −60(___blank)2+ 240(___blank) − 80

Answer: 2, 2

P = −60(4) + 480 − 80

P = −240 + 480 − 80

P = $ ___blank

Answer: 160

So, P = 160 when x = 2. This means the maximum profit will be $160.

2. How much should each bag of cotton candy cost to generate the maximum profit?

You found that P = 160 (the maximum profit) when x = 2. If the x-axis represents the cost per bag, what should the price of each bag be to have the maximum profit?

Cost = $ ___blank per bag

Answer: 2

3. How much profit would be generated if the cotton candy sold for $1.50?

Remember the x-axis represents the cost per bag. Therefore, substitute x = $1.50 and simplify to find P, the profit.

Substitute x = $1.50 into the original equation.

P = −60x2+ 240x − 80

P = −60(___blank)2+ 240(___blank)− 80

Answer: 1.50, 1.50

P = −60( ___blank ) +___blank − 80

Answer: 2.25, 360

P = ___blank

Answer: 145

1. What is the maximum daily profit? $160

2. How much should each bag of cotton candy cost to generate the maximum profit? $2

3. How much profit would be generated if the cotton candy sold for $1.50? $145

Practice Problem #2

A ball is thrown into the air. The path of the ball is modeled by the equation h = t2 + 10t where h represents the height (in feet) of the ball after t seconds. The graph of this equation is shown to the right. Use this graph to answer the questions.

A coordinate plane with the x-axis labeled as time in seconds and the y-axis is labeled as height in feet. A parabola is graphed that is opening down with x-intercepts at (0,0) and (10,0), a y-intercept at (0,0), and a vertex at (5,25).

1. What is the maximum height of the ball? How many seconds after being thrown does this occur?

After ___blank seconds, the maximum height of the ball is ___blank feet in the air.

Answer: 5, 25

A coordinate plane with the x-axis labeled as time in seconds and the y-axis is labeled as height in feet. A parabola is graphed that is opening down with x-intercepts at (0,0) and (10,0), a y-intercept at (0,0), and a vertex at (5,25). There is a point plotted on the vertex at (5,125).

2. How long does it take for the ball to hit the ground?

After ___blank seconds, the ball will reach the ground. This occurs when h = 0.

Answer: 10

A coordinate plane with the x-axis labeled as time in seconds and the y-axis is labeled as height in feet. A parabola is graphed that is opening down with x-intercepts at (0,0) and (10,0), a y-intercept at (0,0), and a vertex at (5,25). There is a point plotted on the x-intercept at (10,0).

3. What will be the height of the ball after 3 seconds?

After 3 seconds, the height of the ball will be ___blank feet.

Answer: 21

A coordinate plane with the x-axis labeled as time in seconds and the y-axis is labeled as height in feet. A parabola is graphed that is opening down with x-intercepts at (0,0) and (10,0), a y-intercept at (0,0), and a vertex at (5,25). There is a point plotted on the parabola at (3,21).

Practice Problem #3

A farmer has 100 feet of fencing to build a rectangular pen, using his barn as one side of the enclosure. What dimensions will maximize the area of the pen? What is the maximum area that can be enclosed?

If x-represents the width of the pen, then 100 − 2x represents the length.

The formula for the area is A = lw where l is the length and w is the width.

l = 100 − 2x

w = x

Using this information, you can substitute into the formula, A = lw.

You will have:

A = lw

A = (100 − 2x)(x)

Simplify using the distributive property.

A = lw

A = (100 − 2x)(x)

A = ___blankx − ___blankx2

Answer: 100, 2

The function A = 100x − 2x2 will give you the maximum area for this pen. Rewrite this equation in standard form to A = −2x2 + 100x.

To find the maximum value, find the y-value of the vertex. Substitute 100 for a and −2 for b into the formula for finding the maximum value x = negative b over 2 a..

A picture of a smaller rectangle below a longer rectangle. The top rectangle is labeled barn. The bottom rectangle is labeled Pen for Animals with the left and right side labeled x and the bottom side labeled 100 minus 2x.

Simplify.

x = negative b over 2 a. = negative the fraction with numerator 100 and denominator 2 times negative 2 equals negative 100 over negative 4 = ____

Answer: 25

This will give you x − 25, which is the width. Substitute to find the length of the fence.

100 − 2x

100 − 2(25)

100 − ___blank

Answer: 50

100 − 50 = ___blank

Answer: 50

You have w = 25 feet and l = 50 feet. Use this to find the area using the formula A = lw.

A = lw

A = (25)(50)

A = ___blank__

Answer: 1250

The corral maximum area has width 25 feet, length 50 feet, and area 1250 ft2.