Deck 4: Polynomial and Rational Functions Applications to Optimization

Question 1

Find all fixed points of the function. $f ( x ) = 13 + \sqrt { x - 1 }$

A) x = 17
B) x = 0, x = 170
C) x = 10
D) x = 17, x = 10
E) no fixed points

Answer 1

x = 17

Question 2

Suppose that the revenue generated by selling x units of a certain commodity is given by $R = - \frac { 1 } { 5 } x ^ { 2 } + 600 x$ . Assume that R is in dollars. What is the maximum revenue possible in this situation?

A) $450,000
B) $440,000
C) $430,000
D) $900,000
E) $480,000

Answer 2

$450,000

Question 3

Sketch the graph of the function and specify all x- and y-intercepts. y = - 3x ⁴ + 5

A) x-intercepts: 5
Y-intercept: - 50,625
B) x-intercepts: -5
Y-intercept: - 50,625
C) x-intercepts: $- \sqrt [ 4 ] { \frac { 5 } { 3 } } , \sqrt [ 4 ] { \frac { 5 } { 3 } }$
Y-intercept: - 5
D) x-intercepts: $- \sqrt [ 4 ] { \frac { 5 } { 3 } } , \sqrt [ 4 ] { \frac { 5 } { 3 } }$
Y-intercept: 5
E) x-intercept: 0
Y-intercept: 0

Answer 3

x-intercepts: $- \sqrt [ 4 ] { \frac { 5 } { 3 } } , \sqrt [ 4 ] { \frac { 5 } { 3 } }$
Y-intercept: 5

Question 4

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$

A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept: $\pm 2$ ; s-intercept: 1.
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept: $\pm \sqrt { 8 }$ ;s-intercept: 2.

Question 5

Sketch the graph of the function and specify all x- and y-intercepts. y =- ( x - 3) ³ - 1

A) x-intercept: -1
Y-intercept: -1
B) x-intercept: -2
Y-intercept: 26
C) x-intercept: 2
Y-intercept: 26
D) x-intercept: 2
Y-intercept: - 26
E) x-intercept: 3
Y-intercept: 27

Question 6

A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ . Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.

A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ .
The square of the area of the triangle is equal to $( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ , and the substitution $x ^ { 2 } = t$ will transform this expression into the quadratic function $- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )$
Since we want to find the maximum value of t, we will substitute the value $t = 18 R ^ { 2 } = x ^ { 2 }$ into the equation. Solving for t gives us the following minimum area of the shaded region: $t = \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ .
The square of the area of the triangle is equal to $( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ The substitution $x ^ { 2 } = t$ will transform this expression into the quadratic function $- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 18 R ^ { 2 }$
Substituting the value $t = 18 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 18 R ^ { 2 }$ and consequently $x = 3 R \sqrt { 2 }$ (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = 3 R \sqrt { 2 }$ in the equation (2) gives us that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to $\frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 }$ .
The substitution $x ^ { 2 } = t$ will transform this into the quadratic function $\frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 }$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 9 R ^ { 2 }$
Substituting the value $t = 9 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 9 R ^ { 2 }$ and consequently $x = 3 R$ . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = 3 R$ into the equation (2), we find that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to $A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }$ , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: $x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 }$
We can then write $x ^ { 2 } = 18 R ^ { 2 }$ as $x ^ { 2 } = 18 R ^ { 2 }$ and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .
E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to $\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }$ .
The substitution $x ^ { 2 } = t$ will transform this into the quadratic function $\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 }$
Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is $t = \frac { - b } { 2 a } = 3 R ^ { 2 }$
Substituting the value $t = 3 R ^ { 2 }$ into the equation $t = x ^ { 2 }$ gives us $x ^ { 2 } = 3 R ^ { 2 }$ and consequently $x = R \sqrt { 7 }$ . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to $\frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } }$
Substituting the value $x = R \sqrt { 7 }$ into the equation (2), we find that the minimum value of the shaded region is equal to $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ .

Question 7

Plot the following points: (6, 2), (7, 5), (8, 8), (9, 9). In your scatter diagram, sketch a line that best seems to fit the data. Estimate the slope and the y-intercept of the line.

A) Estimated slope: 2; Estimated y-intercept: -12.5
B) Estimated slope: 2.5; Estimated y-intercept: -13
C) Estimated slope: 2.9; Estimated y-intercept: -15.3
D) Estimated slope: 1.2; Estimated y-intercept: -1.5
E) Estimated slope: 2.5; Estimated y-intercept: -16

Question 8

Sketch the graph of the function and specify all x- and y-intercepts. y =(x - 2) ² + 4

A) There is no x-intercept. y-intercept: 8
B) There is no x-intercept. y-intercept: 8
C) There is no x-intercept. y-intercept: 7
D) x-intercept: 0 y-intercept: 4
E) There is no x-intercept. y-intercept: 7

Question 9

What is the largest possible area for a rectangle with a perimeter of 40 cm?

A) 400 cm ²
B) 100 cm ²
C) 90 cm ²
D) 130 cm ²
E) 150 cm ²

Question 10

Find the linear function satisfying the given conditions. $g ( 0 ) = 0$ and $g ( 2 ) = \sqrt { 3 }$

A) $g ( x ) = \frac { \sqrt { 3 x } } { 2 }$
B) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x$
C) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x - 3$
D) $g ( x ) = \frac { \sqrt { 3 } } { 2 } x + 2$
E) $g ( x ) = \frac { 2 \sqrt { 3 } } { 3 } x$

Question 11

For the following figure, express the length AB as a function of x. (Hint: Note the similar triangles.)

A) $A B ( x ) = \frac { ( x + 2 ) \sqrt { x ^ { 2 } + 9 } } { x ^ { 2 } }$
B) $A B ( x ) = \frac { ( x + 2 ) \sqrt { x ^ { 2 } + 9 } } { x }$
C) $A B ( x ) = \frac { ( x + 3 ) \sqrt { x + 4 } } { x }$
D) $A B ( x ) = \frac { ( x + 3 ) ^ { 2 } ( x + 2 ) } { x }$
E) $A B ( x ) = \frac { ( x + 3 ) \left( x ^ { 2 } + 2 \right) } { x }$

Question 12

Find all fixed points of the function. $f ( x ) = - 3 x + 4$

A) x = 1
B) x = - 1
C) x = 4
D) x = 3
E) no fixed points

Question 13

A factory owner buys a new machine for $25,000. After eight years, the machine has a salvage value of $1,000. Find a formula for the value of the machine after t years, where $0 \leq t \leq 8$

A) V (t) = - 3,000x + 25,000
B) V (t) = - 3,000x + 1,000
C) V (t) = 3,000x + 25,000
D) V (t) = 3,000x - 25,000
E) V (t) = - 3,000x - 25,000

Question 14

Sketch the graph of the function and specify all x- and y-intercepts. y = (x - 2)(x - 1)(x + 1)

A) x-intercepts: - 2, 1, 2
Y-intercept: -4
B) x-intercepts: - 1, 1, 2
Y-intercept: 2
C) x-intercepts: - 1, 1, 2
Y-intercept: -2
D) x-intercepts: - 2, 1, 2
Y-intercept: 4
E) x-intercepts: - 1, 1
Y-intercepts: - 2.25

Question 15

Two points A and B move along the x-axis. After t sec, their positions are given by the equations A: $\quad x = 4 t + 120$
B: $\quad x = 30 t - 36$ At what time t do A and B have the same x-coordinate?

A) 7 sec
B) 6 sec
C) 5 sec

Question 16

Find all fixed points of the function. $h ( x ) = x ^ { 2 } - 7 x - 9$

A) x = - 1, x = - 9
B) x = - 1, x = 9
C) x = - 9, x = 1
D) x = 9
E) no fixed points

Question 17

Determine the input that produces the largest or smallest output (whichever is appropriate). State whether the output is largest or smallest. s = - 24t ² + 147t + 15

A) $t = \frac { 49 } { 16 }$ ; largest
B) $t = \frac { 16 } { 49 }$ ; largest
C) $x = 49$ ; smallest
D) $x = 49$ ; largest
E) $t = \frac { 49 } { 16 }$ ; smallest

Question 18

A piece of wire $7 \pi y$ inches long is bent into a circle. Express the area of the circle as a function of y.

A) $A ( x ) = 49 \pi y ^ { 2 }$
B) $A ( x ) = \frac { 49 \pi y ^ { 2 } } { 2 }$
C) $A ( x ) = 25 y ^ { 2 }$
D) $A ( x ) = \frac { 25 \pi y ^ { 2 } } { 4 }$
E) $A ( x ) = \frac { 49 \pi y ^ { 2 } } { 4 }$

Question 19

Suppose that the height of an object shot straight up is given by h =544t - 16t ² . (Here h is in feet and t is in seconds.) Find the maximum height.

A) 4,724 ft
B) 4,624 ft
C) 1,156 ft
D) 4,290 ft
E) 5,189 ft

Question 20

Among all rectangles having a perimeter of 24m, find the dimensions of the one with the largest area.

A) $9 \mathrm {~m} \text { by } 15 \mathrm {~m}$
B) $\frac { 9 } { 4 } \mathrm {~m} \text { by } \frac { 15 } { 4 } \mathrm {~m}$
C) $4 \mathrm {~m} \text { by } 6 \mathrm {~m}$
D) $6 \mathrm { mby } 6 \mathrm {~m}$
E) $\frac { 9 } { 2 } \mathrm {~m} \text { by } \frac { 15 } { 2 } \mathrm {~m}$

Question 21

Sketch the graph of the rational function. Specify the intercepts and the asymptotes. $y = \frac { 1 } { ( x + 1 ) ^ { 2 } }$

A) no x-intercepts; y-intercept: 1; no vertical asymptotes; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: 1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: 1; vertical asymptote: x = - 1; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: -1; vertical asymptote: x = 1; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: -1; no vertical asymptotes; horizontal asymptote: y = 0;

Question 22

Sketch the graph of the function and specify all x- and y-intercepts.y = x ³ - 9x

A) x-intercept: 0
Y-intercept: 0
B) x-intercept: 0
Y-intercept: 0
C) x-intercepts: -3, 0, 3
Y-intercept: 0
D) x-intercepts: -3, 0, 3
Y-intercept: 0
E) x-intercept: 0
Y-intercept: 0

Question 23

Sketch the graph of the rational function. Specify the intercepts and the asymptotes. $y = \frac { x - 3 } { x + 3 }$

A) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = - 3; horizontal asymptote: y = 1;
B) x-intercept: 0; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
C) x-intercept: 3; y-intercept: 1; vertical asymptote: x = - 3; horizontal asymptote: y = - 1;
D) x-intercept: 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = - 1;
E) x-intercept: - 3; y-intercept: - 1; vertical asymptote: x = 3; horizontal asymptote: y = 1;

Question 24

Sketch the graph of the function and specify all x- and y-intercepts. y = x ³ (x + 1)

A) x-intercepts: 0, 1
Y-intercept: 0
B) x-intercepts: 0, - 1
Y-intercept: 0
C) x-intercepts: 0, 1
Y-intercept: 0
D) x-intercepts: 0, 1
Y-intercept: 0
E) x-intercepts: 0, - 1
Y-intercept: 0

Question 25

Sketch the graph of the rational function. Specify the intercepts and the asymptotes. $y = - \frac { 3 } { x - 4 }$

A) no x-intercepts; y-intercept: $- \frac { 3 } { 4 }$ ; vertical asymptote: x = - 4; horizontal asymptote: y = 0;
B) no x-intercepts; y-intercept: $\frac { 4 } { 3 }$ ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
C) no x-intercepts; y-intercept: $- \frac { 4 } { 3 }$ ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
D) no x-intercepts; y-intercept: $\frac { 3 } { 4 }$ ; vertical asymptote: x = 4; horizontal asymptote: y = 0;
E) no x-intercepts; y-intercept: $\frac { 1 } { 4 }$ ; vertical asymptote: x = 4; horizontal asymptote: y = 0;