Question 1

Solve the problem.
-A wire 20 feet long is to be cut into two pieces. One piece will be shaped as a square and the other piece will be
shaped as an equilateral triangle. Express the total area A enclosed by the pieces of wire as a function of the
length x of a side of the equilateral triangle. What is the domain of A?

Accepted Answer

The answer of Solve the problem.
-A wire 20 feet long...

Question 2

Solve the problem.
-Two boats leave a dock at the same time. One boat is headed directly east at a constant speed of 35 knots
(nautical miles per hour), and the other is headed directly south at a constant speed of 22 knots. Express the
distance d between the boats as a function of the time t.

Accepted Answer

The answer of Solve the problem.
-Two boats leave a dock...

Question 3

Solve the problem.
-Bob wants to fence in a rectangular garden in his yard. He has 64 feet of fencing to work with and wants to use it all. If the garden is to be x feet wide, express the area of the garden as a function of x.

Accepted Answer

The perimeter of the rectangle is given as $2x + 2y = 64$, where $x$ is the width and $y$ is the length. Solving for $y$, we get $y = 32 - x$. The area $A$ of the rectangle is $xy$, substituting $y$ with $32 - x$, we get $A(x) = x(32 - x) = 32x - x^2$.

Question 4

Solve the problem.
-A rectangle that is x feet wide is inscribed in a circle of radius 36 feet. Express the area of the rectangle as a function of x.

Accepted Answer

The diameter of the circle is the diagonal of the rectangle, which is $2 \times 36 = 72$ feet. By the Pythagorean theorem, if $x$ is the width, then the height $h$ of the rectangle can be found from $x^2 + h^2 = 72^2$. Solving for $h$, we get $h = \sqrt{72^2 - x^2}$. The area of the rectangle is $A(x) = x \cdot h = x \cdot \sqrt{72^2 - x^2} = x \cdot \sqrt{5184 - x^2}$.

Question 5

Solve the problem.
-From a 48-inch by 48-inch piece of metal, squares are cut out of the four corners so that the sides can then be folded up to make a box. Let x represent the length of the sides of the squares, in inches, that are cut out. Express
The volume of the box as a function of x.

Accepted Answer

After cutting out squares of side length $x$ from each corner, the dimensions of the box will be $(48-2x)$ by $(48-2x)$ by $x$, since the height of the box will be $x$ and the length and width will each be reduced by $2x$ (once from each side). The volume $V$ of the box is the product of its dimensions, so $V(x) = x(48-2x)(48-2x)$. Expanding this gives $V(x) = x(2304 - 192x + 4x^2) = 4x^3 - 192x^2 + 2304x$.

Question 6

Solve the problem.
-The price $p$ and $x$, the quantity of a certain product sold, obey the demand equation
$$p = - \frac { 1 } { 10 } x + 100 , \{ x \mid 0 \leq x \leq 1000 \}$$
a) Express the revenue $R$ as a function of $x$.
b) What is the revenue if 450 units are sold?
c) Graph the revenue function using a graphing utility.
d) What quantity x maximizes revenue? What is the maximum revenue?
e) What price should the company charge to maximize revenue?

Accepted Answer

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

Question 7

Solve the problem.
-Sue wants to put a rectangular garden on her property using 76 meters of fencing. There is a river that runs through her property so she decides to increase the size of the garden by using the river as one side of the
Rectangle. (Fencing is then needed only on the other three sides.) Let x represent the length of the side of the
Rectangle along the river. Express the garden's area as a function of x.

Accepted Answer

The perimeter of the garden is 76 meters, so the other three sides have a total length of 76 - x meters. Let y be the width of the garden. Then the perimeter is:$$x + 2y = 76 - x$$$$3x + 2y = 76$$$$2y = 76 - 3x$$$$y = 38 - \frac{3}{2}x$$The area of the garden is:$$A = xy = x(38 - \frac{3}{2}x) = 38x - \frac{3}{2}x^2$$

Question 8

Solve the problem.
-A farmer has 1,600 yards of fencing to enclose a rectangular garden. Express the area A of the rectangle as a function of the width x of the rectangle. What is the domain of A?

Accepted Answer

The perimeter of the rectangle is given as 1,600 yards, which can be expressed as $2x + 2y = 1,600$, where $x$ is the width and $y$ is the length. Solving for $y$, we get $y = 800 - x$. The area $A$ of the rectangle is $x \times y$, which gives $A(x) = x(800 - x) = 800x - x^2$. The domain of $A$ is determined by the physical constraints of the problem: $x$ must be greater than 0 but less than 800, because if $x$ is 800, then $y$ would be 0 (resulting in no area), and $x$ cannot be more than 800 because it would make $y$ negative, which is not possible for the dimensions of a rectangle.

Question 9

Solve the problem.
-A rectangular box with volume 433 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 per square foot for the sides. Let x represent the length of a side of the
Base. Express the cost the box as a function of x.

Accepted Answer

The cost function $C(x)$ combines the cost of the top and bottom, which is $2 \times 1.50x^2 = 3x^2$, with the cost of the sides. The volume is $x^2h = 433$, so $h = \frac{433}{x^2}$. The area of the four sides is $4xh = 4x \times \frac{433}{x^2} = \frac{1732}{x}$, and at $2.00 per square foot, the cost for the sides is $2 \times \frac{1732}{x} = \frac{3464}{x}$. Thus, $C(x) = 3x^2 + \frac{3464}{x}$.

Question 10

Solve the problem.
-The figure shown here shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 4 units long. Express the area A of the rectangle in terms of x.

Accepted Answer

The answer of Solve the problem.
-The figure shown here shows...

Question 11

Solve the problem.
-A rocket is shot straight up in the air from the ground at a rate of 80 feet per second. The rocket is tracked by a range finder that is 431 feet from the launch pad. Let d represent the distance from the rocket to the range finder
And t represent the time, in seconds, since "blastoff". Express d as a function of t.

Accepted Answer

The distance $d(t)$ from the rocket to the range finder can be found using the Pythagorean theorem, where one leg is the constant distance of 431 feet from the launch pad to the range finder, and the other leg is the vertical distance the rocket has traveled, $80t$. Thus, $d(t) = \sqrt{431^2 + (80t)^2}$.

Question 12

Solve the problem.
-A wire of length 3x is bent into the shape of a square. Express the area A of the square as a function of x.

Accepted Answer

The perimeter of the square is 3x, so each side is $\frac{3x}{4}$. The area is $(\frac{3x}{4})^2 = \frac{9}{16}x^2$.

Question 13

Solve the problem.
-A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 11 inches by 24 inches by cutting out equal squares of side x at each corner and then folding up the sides as in the figure.
Express the volume V of the box as a function of x.

Accepted Answer

The answer of Solve the problem.
-A box with an open...

Question 14

Solve the problem.
-Let $P = ( x , y )$ be a point on the graph of $y = \sqrt { x }$. Express the distance $d$ from $P$ to the point $( 1,0 )$ as a function of $x$

Accepted Answer

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Here, $P = (x, \sqrt{x})$ and the point is $(1, 0)$, so $d(x) = \sqrt{(x - 1)^2 + (\sqrt{x} - 0)^2} = \sqrt{x^2 - 2x + 1 + x} = \sqrt{x^2 - x + 1}$.

Question 15

Solve the problem.
-A right triangle has one vertex on the graph of $$y = x ^ { 2 }$$ at (x, y), another at the origin, and the third on the
(positive) y-axis at (0, y). Express the area A of the triangle as a function of x.

Accepted Answer

The answer of Solve the problem.
-A right triangle has one...

Question 16

Solve the problem.
-The volume V of a square-based pyramid with base sides s and height h is $V = \frac { 1 } { 3 } s ^ { 2 } h$ If the height is half of the
length of a base side, express the volume V as a function of s.

Accepted Answer

The answer of Solve the problem.
-The volume V of a...

Question 17

Solve the problem.
-A farmer's silo is the shape of a cylinder with a hemisphere as the roof. If the height of the silo is 66 feet and the radius of the hemisphere is r feet, express the volume of the silo as a function of r.

Accepted Answer

The volume of the silo is the sum of the volume of the cylinder and the volume of the hemisphere. The volume of the cylinder is given by $V_{cylinder} = \pi r^2 h$, where $h = 66 - r$ (since the total height is 66 feet and the hemisphere's height is $r$, so the height of the cylindrical part is $66 - r$). The volume of the hemisphere is given by $V_{hemisphere} = \frac{2}{3}\pi r^3$. Therefore, the total volume $V(r)$ is $V(r) = \pi (66 - r)r^2 + \frac{2}{3}\pi r^3$.

Question 18

Solve the problem.
-The price p and the quantity x sold of a certain product obey the demand equation: $$p = - \frac { 1 } { 5 } x + 200 , \{ x \mid 0 \leq x \leq 500 \}$$ What is the revenue to the nearest dollar when 300 units are sold?

Accepted Answer

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

Question 19

Solve the problem.
-A farmer's silo is the shape of a cylinder with a hemisphere as the roof. If the radius of the hemisphere is 10 feet and the height of the silo is h feet, express the volume of the silo as a function of h.

Accepted Answer

The volume of the silo is the sum of the volume of the cylinder and the hemisphere. The volume of the cylinder is $V_{cylinder} = \pi r^2 h$, where $r = 10$ feet and $h$ is the height of the cylindrical part. The volume of the hemisphere is $V_{hemisphere} = \frac{2}{3} \pi r^3$, with $r = 10$ feet. Thus, the total volume is $V(h) = \pi (10)^2 h + \frac{2}{3} \pi (10)^3 = 100\pi h + \frac{2000}{3}\pi$. Note that $h$ in the volume formula for the cylinder part does not include the hemisphere's height, as it represents only the height of the cylindrical part.

Question 20

Solve the problem.
-A rectangular sign is being designed so that the length of its base, in feet, is 10 feet less than 4 times the height, h. Express the area of the sign as a function of h.

Accepted Answer

The area of a rectangle is given by the formula $A = l \times w$, where $l$ is the length and $w$ is the width (or height in this case). Given that the length is $10$ feet less than $4$ times the height, we can express the length as $l = 4h - 10$. Therefore, the area as a function of height $h$ is $A(h) = h \times (4h - 10) = 4h^2 - 10h$.

Quiz 1: Functions and Their Graphs

Solve the problem. -Two boats leave a dock at the same time. One boat is headed directly east at a constant speed of 35 knots (nautical miles per hour), and the other is headed directly south at a constant speed of 22 knots. Express the distance d between the boats as a function of the time t.

Solve the problem. -Bob wants to fence in a rectangular garden in his yard. He has 64 feet of fencing to work with and wants to use it all. If the garden is to be x feet wide, express the area of the garden as a function of x.

Solve the problem. -A rectangle that is x feet wide is inscribed in a circle of radius 36 feet. Express the area of the rectangle as a function of x.

Solve the problem. -From a 48-inch by 48-inch piece of metal, squares are cut out of the four corners so that the sides can then be folded up to make a box. Let x represent the length of the sides of the squares, in inches, that are cut out. Express The volume of the box as a function of x.

Solve the problem. -A farmer has 1,600 yards of fencing to enclose a rectangular garden. Express the area A of the rectangle as a function of the width x of the rectangle. What is the domain of A?

Solve the problem. -A rectangular box with volume 433 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 per square foot for the sides. Let x represent the length of a side of the Base. Express the cost the box as a function of x.

Solve the problem. -A wire of length 3x is bent into the shape of a square. Express the area A of the square as a function of x.

Solve the problem. -A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 11 inches by 24 inches by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a function of x.

Solve the problem. -Let $P = ( x , y )$ be a point on the graph of $y = \sqrt { x }$ . Express the distance $d$ from $P$ to the point $( 1,0 )$ as a function of $x$

Solve the problem. -A right triangle has one vertex on the graph of $y = x ^ { 2 }$ at (x, y), another at the origin, and the third on the (positive) y-axis at (0, y). Express the area A of the triangle as a function of x.

Solve the problem. -The volume V of a square-based pyramid with base sides s and height h is $V = \frac { 1 } { 3 } s ^ { 2 } h$ If the height is half of the length of a base side, express the volume V as a function of s.

Solve the problem. -A farmer's silo is the shape of a cylinder with a hemisphere as the roof. If the height of the silo is 66 feet and the radius of the hemisphere is r feet, express the volume of the silo as a function of r.

Solve the problem. -The price p and the quantity x sold of a certain product obey the demand equation: $p = - \frac { 1 } { 5 } x + 200 , \{ x \mid 0 \leq x \leq 500 \}$ What is the revenue to the nearest dollar when 300 units are sold?

Solve the problem. -A farmer's silo is the shape of a cylinder with a hemisphere as the roof. If the radius of the hemisphere is 10 feet and the height of the silo is h feet, express the volume of the silo as a function of h.

Solve the problem. -A rectangular sign is being designed so that the length of its base, in feet, is 10 feet less than 4 times the height, h. Express the area of the sign as a function of h.