Question 1

Solve the problem.
-You have 348 feet of fencing to enclose a rectangular region. What is the maximum area?

Accepted Answer

The maximum area that can be enclosed by a given length of fencing is achieved by forming a square. The perimeter of the square is 348 feet, so each side is 348/4 = 87 feet. The area of the square is 87^2 = 7,569 square feet.

Question 2

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then
find that value.
-$$f ( x ) = - 10 x ^ { 2 } - 2 x - 8$$

Accepted Answer

The quadratic function has a maximum value because the coefficient of $x^2$ is negative. The vertex form gives the maximum value at the vertex, calculated as $-\frac{b^2 - 4ac}{4a}$, which results in $-\frac{79}{10}$.

Question 3

Solve the problem.
-You have 336 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area.

Accepted Answer

To maximize the area of a rectangle with a fixed perimeter, the rectangle should be a square. With a perimeter of 336 feet, each side of the square would be 336/4 = 84 feet.

Question 4

Solve the problem.
-The price $p$ and the quantity $x$ sold of a certain product obey the demand equation
$$p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200 \text {. }$$
What quantity $x$ maximizes revenue? What is the maximum revenue?

Accepted Answer

Revenue $R$ is given by the product of price $p$ and quantity $x$, so $R = px$. Substituting the given demand equation into this formula gives $R = (-\frac{1}{6}x + 200)x = -\frac{1}{6}x^2 + 200x$. To maximize revenue, we take the derivative of $R$ with respect to $x$ and set it to zero: $\frac{dR}{dx} = -\frac{1}{3}x + 200 = 0$, solving for $x$ gives $x = 600$. Substituting $x = 600$ back into the revenue equation gives the maximum revenue: $R = -\frac{1}{6}(600)^2 + 200(600) = 60,000$.

Question 5

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then
find that value.
-$$f ( x ) = 3 x ^ { 2 } + 2 x - 6$$

Accepted Answer

Since the coefficient of $x^2$ is positive (3), the parabola opens upwards, indicating a minimum value. To find the minimum value, use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Here, $a = 3$ and $b = 2$, so $x = -\frac{2}{2*3} = -\frac{1}{3}$. Substitute $x = -\frac{1}{3}$ into the original equation to find the y-coordinate (minimum value): $f(-\frac{1}{3}) = 3(-\frac{1}{3})^2 + 2(-\frac{1}{3}) - 6 = 3(\frac{1}{9}) - \frac{2}{3} - 6 = \frac{1}{3} - \frac{2}{3} - 6 = -\frac{1}{3} - 6 = -\frac{19}{3}$.

Question 6

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then
find that value.
-$$f ( x ) = - 5 x ^ { 2 } + 5 x$$

Accepted Answer

The answer of Determine, without graphing, whether the given quadratic...

Question 7

Solve the problem.
-Consider the quadratic model h(t) = -16t2 + 40t + 50 for the height (in feet), h, of an object t seconds after the object has been projected straight up into the air. Find the maximum height attained by the object. How much
Time does it take to fall back to the ground? Assume that it takes the same time for going up and coming down.

Accepted Answer

The maximum height can be found by using the vertex formula for a quadratic equation, $ h(t) = -16t^2 + 40t + 50 $. The time at the vertex, $ t_v $, is given by $ t_v = -\frac{b}{2a} = -\frac{40}{2(-16)} = 1.25 $ seconds. Plugging $ t_v $ back into the equation gives the maximum height, $ h(1.25) = -16(1.25)^2 + 40(1.25) + 50 = 75 $ feet. To find the time it takes to hit the ground, set $ h(t) = 0 $ and solve for $ t $, which gives two solutions, one of which is $ t = 2.5 $ seconds (the other solution is the start time, $ t = 0 $). This accounts for the time up and down, confirming the object takes 2.5 seconds to hit the ground after being projected.

Question 8

Solve the problem.
-An object is propelled vertically upward from the top of a 128-foot building. The quadratic function models the ball's height above the ground, s(t), in feet, t seconds after it was thrown. $$s ( t ) = - 16 t ^ { 2 } + 112 t + 128$$ How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if
Necessary.

Accepted Answer

To find when the object hits the ground, set $s(t) = 0$ and solve for $t$. The equation becomes $-16t^2 + 112t + 128 = 0$. Factoring or using the quadratic formula, we find the time $t$ when the object hits the ground. The correct answer, when solved, is approximately 8.0 seconds.

Question 9

Solve the problem.
-The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $x = - 6 p + 120 , \quad 0 \leq p \leq 20$.
What quantity $x$ maximizes revenue? What is the maximum revenue?

Accepted Answer

Revenue $R$ is given by $R = xp = (-6p + 120)p$. To maximize revenue, differentiate $R$ with respect to $p$ and set the derivative equal to zero. The derivative $R' = -12p + 120$. Setting $R' = 0$ gives $p = 10$. Substituting $p = 10$ into the demand equation gives $x = -6(10) + 120 = 60$. The maximum revenue is $R = 10 \times 60 = \$600$.

Question 10

Solve the problem.
-The owner of a video store has determined that the cost $\mathrm { C }$, in dollars, of operating the store is approximately given by $C ( x ) = 2 x ^ { 2 } - 32 x + 530$, where $x$ is the number of videos rented daily. Find the lowest cost to the nearest dollar.

Accepted Answer

The cost function $C(x) = 2x^2 - 32x + 530$ is a quadratic function that opens upwards (since the coefficient of $x^2$ is positive). The vertex of this parabola represents the minimum point. The x-coordinate of the vertex is given by $-b/(2a)$, where $a = 2$ and $b = -32$. Calculating this gives $x = 8$. Substituting $x = 8$ back into the cost function gives the minimum cost: $C(8) = 2(8)^2 - 32(8) + 530 = 402$. Thus, the lowest cost is $\$402$.

Question 11

Solve the problem.
-A projectile is fired from a cliff 300 feet above the water at an inclination of $45 ^ { \circ }$ to the horizontal, with a muzzle velocity of 270 feet per second. The height $h$ of the projectile above the water is given by $h ( x ) = \frac { - 32 x ^ { 2 } } { ( 270 ) ^ { 2 } } + x + 300$, where $\mathrm { x }$ is the horizontal distance of the projectile from the base of the cliff. Find the maximum height of the projectile.

Accepted Answer

The maximum height of the projectile can be found by completing the square or using the vertex formula for a parabola. The given equation is in the form of $h(x) = ax^2 + bx + c$, where $a = \frac{-32}{(270)^2}$, $b = 1$, and $c = 300$. The vertex of this parabola, which gives the maximum height, occurs at $x = -\frac{b}{2a}$. Substituting the values of $a$ and $b$ into this formula and then calculating $h(x)$ at this $x$ value will give the maximum height, which is $869.53$ feet.

Question 12

Solve the problem.
-The price $p$ and the quantity $x$ sold of a certain product obey the demand equation
$$p = - \frac { 1 } { 4 } x + 120,0 \leq x \leq 480 .$$
What price should the company charge to maximize revenue?

Accepted Answer

The answer of Solve the problem.
-The price $p$ and the...

Question 13

Solve the problem.
-The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation
$$p = - 10 x + 240 , \quad 0 \leq x \leq 24 \text {. }$$
What price should the company charge to maximize revenue?

Accepted Answer

Revenue $R$ is given by $R = p \times x = (-10x + 240)x$. To maximize revenue, differentiate $R$ with respect to $x$ and set the derivative equal to zero. The derivative of $R$ with respect to $x$ is $-20x + 240$. Setting this equal to zero gives $x = 12$. Substituting $x = 12$ into the demand equation gives $p = -10(12) + 240 = 120$, which means the price should be \$12 to maximize revenue.

Question 14

Solve the problem.
-A projectile is fired from a cliff 300 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 300 feet per second. The height h of the projectile above the water is given by $$h ( x ) = \frac { - 32 x ^ { 2 } } { ( 300 ) ^ { 2 } } + x + 300$$ where x is the horizontal distance of the projectile from the base of the cliff. How far from the base of the cliff is
The height of the projectile a maximum?

Accepted Answer

The maximum height of a projectile in a parabolic trajectory occurs at the vertex of the parabola. The x-coordinate of the vertex for a quadratic equation $ax^2 + bx + c$ is given by $-b/(2a)$. Here, $a = -32/(300)^2$ and $b = 1$. Calculating $-b/(2a)$ gives approximately 1,406.25 ft.

Question 15

Solve the problem.
-A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 276 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

Accepted Answer

The largest area that can be enclosed by a given perimeter is a square. However, since one side is not fenced, the optimal shape becomes a rectangle where the length is twice the width. The total fencing available is 276 feet, which will be used for two widths and one length. Let the width be $w$ and the length be $2w$. The perimeter equation is $2w + 2w = 276$, simplifying to $4w = 276$, so $w = 69$ feet. The length is $2w = 138$ feet. The area is $w \times 2w = 69 \times 138 = 9,522$ square feet.

Question 16

Solve the problem.
-The profit that the vendor makes per day by selling $x$ pretzels is given by the function $P ( x ) = - 0.002 x ^ { 2 } + 1.6 x - 350$. Find the number of pretzels that must be sold to maximize profit.

Accepted Answer

The profit function is a quadratic equation in the form $ax^2 + bx + c$, where $a = -0.002$, $b = 1.6$, and $c = -350$. The maximum profit occurs at the vertex of the parabola, which is given by $x = -\frac{b}{2a}$. Substituting the values, $x = -\frac{1.6}{2(-0.002)} = 400$.

Question 17

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then
find that value.
-$$f ( x ) = 4 x ^ { 2 } + 12 x$$

Accepted Answer

The quadratic function is in the form $f(x) = ax^2 + bx + c$, where $a = 4$, $b = 12$, and $c = 0$. Since $a > 0$, the parabola opens upwards, indicating a minimum value. To find the minimum value, use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. Substituting $a = 4$ and $b = 12$ gives $x = -\frac{12}{2(4)} = -\frac{12}{8} = -\frac{3}{2}$. Then, substitute $x = -\frac{3}{2}$ into the original equation to find the minimum value: $f(-\frac{3}{2}) = 4(-\frac{3}{2})^2 + 12(-\frac{3}{2}) = 4(\frac{9}{4}) - 18 = 9 - 18 = -9$.

Question 18

Solve the problem.
-The owner of a video store has determined that the profits P of the store are approximately given by where x is the number of videos rented daily. Find the maximum profit to the nearest $$P ( x ) = - x ^ { 2 } + 150 x + 54$$ dollar.

Accepted Answer

The answer of Solve the problem.
-The owner of a video...

Question 19

Solve the problem.
-You have 64 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area.

Accepted Answer

To maximize the area of a rectangle when one side (the length along the river) is not fenced, you use the fencing for the other three sides. Let the width be $w$ and the length be $l$. The total fencing used is $2w + l = 64$ feet (since the length is only fenced on one side). To maximize the area $A = l \times w$, we express $l$ in terms of $w$ from the fencing constraint: $l = 64 - 2w$. Substituting into the area formula gives $A = w(64 - 2w)$. Differentiating $A$ with respect to $w$ and setting the derivative to zero to find the maximum area, we get $64 - 4w = 0$, solving this gives $w = 16$ feet. Substituting $w = 16$ back into the equation for $l$, we get $l = 64 - 2(16) = 32$ feet. Therefore, the dimensions that maximize the area are a length of 32 feet and a width of 16 feet.

Question 20

Solve the problem.
-The manufacturer of a CD player has found that the revenue R (in dollars) is when the unit price is p dollars. If the manufacturer sets the price p to maximize revenue, $$R ( p ) = - 4 p ^ { 2 } + 1,910 p$$ what is the maximum revenue to the nearest whole dollar?

Accepted Answer

The answer of Solve the problem.
-The manufacturer of a CD...

Quiz 2: Linear and Quadratic Functions

Solve the problem. -You have 348 feet of fencing to enclose a rectangular region. What is the maximum area?

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - $f ( x ) = - 10 x ^ { 2 } - 2 x - 8$

Solve the problem. -You have 336 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area.

Solve the problem. -The price $p$ and the quantity $x$ sold of a certain product obey the demand equation $p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200 \text {. }$ What quantity $x$ maximizes revenue? What is the maximum revenue?

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - $f ( x ) = 3 x ^ { 2 } + 2 x - 6$

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - $f ( x ) = - 5 x ^ { 2 } + 5 x$

Solve the problem. -The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $x = - 6 p + 120 , \quad 0 \leq p \leq 20$ . What quantity $x$ maximizes revenue? What is the maximum revenue?

Solve the problem. -The owner of a video store has determined that the cost $\mathrm { C }$ , in dollars, of operating the store is approximately given by $C ( x ) = 2 x ^ { 2 } - 32 x + 530$ , where $x$ is the number of videos rented daily. Find the lowest cost to the nearest dollar.

Solve the problem. -The price $p$ and the quantity $x$ sold of a certain product obey the demand equation $p = - \frac { 1 } { 4 } x + 120,0 \leq x \leq 480 .$ What price should the company charge to maximize revenue?

Solve the problem. -The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $p = - 10 x + 240 , \quad 0 \leq x \leq 24 \text {. }$ What price should the company charge to maximize revenue?

Solve the problem. -A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 276 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

Solve the problem. -The profit that the vendor makes per day by selling $x$ pretzels is given by the function $P ( x ) = - 0.002 x ^ { 2 } + 1.6 x - 350$ . Find the number of pretzels that must be sold to maximize profit.

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - $f ( x ) = 4 x ^ { 2 } + 12 x$

Solve the problem. -The owner of a video store has determined that the profits P of the store are approximately given by where x is the number of videos rented daily. Find the maximum profit to the nearest $P ( x ) = - x ^ { 2 } + 150 x + 54$ dollar.

Solve the problem. -You have 64 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area.

Solve the problem. -The manufacturer of a CD player has found that the revenue R (in dollars) is when the unit price is p dollars. If the manufacturer sets the price p to maximize revenue, $R ( p ) = - 4 p ^ { 2 } + 1,910 p$ what is the maximum revenue to the nearest whole dollar?

Quiz 2: Linear and Quadratic Functions

Solve the problem. -You have 348 feet of fencing to enclose a rectangular region. What is the maximum area?

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - f(x) =−10x2−2x−8f ( x ) = - 10 x ^ { 2 } - 2 x - 8f(x) =−10x2−2x−8

Solve the problem. -You have 336 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area.

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - f(x) =3x2+2x−6f ( x ) = 3 x ^ { 2 } + 2 x - 6f(x) =3x2+2x−6

Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value. - f(x) =−5x2+5xf ( x ) = - 5 x ^ { 2 } + 5 xf(x) =−5x2+5x

Solve the problem. -The price ppp (in dollars) and the quantity xxx sold of a certain product obey the demand equation x=−6p+120,0≤p≤20x = - 6 p + 120 , \quad 0 \leq p \leq 20x=−6p+120,0≤p≤20 . What quantity xxx maximizes revenue? What is the maximum revenue?

Solve the problem. -The price ppp and the quantity xxx sold of a certain product obey the demand equation p=−14x+120,0≤x≤480.p = - \frac { 1 } { 4 } x + 120,0 \leq x \leq 480 .p=−41​x+120,0≤x≤480. What price should the company charge to maximize revenue?

Solve the problem. -The price ppp (in dollars) and the quantity xxx sold of a certain product obey the demand equation p=−10x+240,0≤x≤24. p = - 10 x + 240 , \quad 0 \leq x \leq 24 \text {. }p=−10x+240,0≤x≤24. What price should the company charge to maximize revenue?