Setting Up Functions

Sol 1

1. The perimeter is given by $P=2x+2y$.

2. $A=xy=50$, so $x=\frac{50}{y}$.

3. Substituting back gives $P=2(50/y) + 2y = 2y +\frac{100}{y}$.

Sol 2

1. The area is given by $A=xy$.

2. The fencing satisfies $F=x+2y=220$, so $x=220-2y$.

3. Substituting back gives $A=(220-2y)y=220y-2y^2$.

Sol 3

1. The area is given by $A=xy$.

2. A diagonal of the rectangle will be a diameter of the circle, so its length is 10. Then $x^2+y^2=10^2=100$ by the Pythagorean Theorem, so $y^2=100-x^2$ and $y=\sqrt{100-x^2}$ since $y>0$.

3. Substituting back gives $A=x\sqrt{100-x^2}$.

Sol 4

1. The area of the rectangle is given by $A=bh=(2x)y$.

2. Since the upper right vertex $(x,y)$ is on the given parabola, we have that $y=9-x^2$.

3. Substituting back gives $A=(2x)(9-x^2)=18x-2x^3$.

Sol 5

1. The area of the triangle is given by $A=\frac{1}{2}bh=\frac{1}{2}xy$.

2. The slope of the hypotenuse is given by

$$m=\frac{y-5}{0-3}=\frac{0-5}{x-3},$$

so solving for $y$ gives

$$y-5=\frac{15}{x-3}$$

so

$$y=\frac{15}{x-3}+5=\frac{5x}{x-3}.$$

3. Substituting back gives

$$A=\frac{1}{2}x\left(\frac{5x}{x-3}\right)$$

$$=\frac{5}{2}\left(\frac{x^2}{x-3}\right)$$

.

Sol 6

1. Since the two semicircular regions can be combined to give a circular region, the area of the field is given by $A=x(2r)+\pi r^2$.

2. Since the perimeter is 400 meters, $P=2x+2\pi r=400$ so $x+\pi r=200$ and $x=200-\pi r$.

3. Substituting back gives $A=(200-\pi r)(2r)+\pi r^2=400r-2\pi r^2+\pi r^2$ $=400r-\pi r^2$.

Sol 7

1. The area of the page is given by $A=bh$.

2 We have that $b=x+2$ and $h=y+3$, and that the area of the printed material is given by $a=xy=50$, so $y=\frac{50}{x}$.

3. Substituting back gives $A=(x+2)(y+3)=(x+2)(\frac{50}{x}+3)$ $=50+\frac{100}{x}+3x+6=3x+\frac{100}{x}+56$.

Sol 8

1. Since the top and bottom each have area given by $x^2$ and the other 4 sides each have area given by $xh$, the total surface area is given by $A=2x^2+4xh$.

2. Since the volume is 80 cubic inches, $V=lwh=x^2h=80$ and therefore $h=\frac{80}{x^2}$.

3. Substituting back gives

$$A=2x^2+4x\left(\frac{80}{x^2}\right)=2x^2+\frac{320}{x}$$

.

Sol 9

1. The cost of the top and bottom is given by $10(2x^2)$, and the cost of the other 4 sides is given by $8(4xh)$, so the total cost is expressed by $C=20x^2+32xh$.

2. Since the volume is 60 cubic inches, $V=lwh=x^2h=60$ and therefore $h=\frac{60}{x^2}$.

3. Substituting back gives $C=20x^2+32x(\frac{60}{x^2})=20x^2+\frac{1920}{x}$.

Sol 10

1. Since the top and the bottom each have area $\pi r^2$, the total surface area is given by $S=2\pi r^2+2\pi rh$.

2. The volume of the cylinder is the area of the base multiplied by the height, so

$V=\pi r^2 h=60\pi$ and $h=\frac{60}{r^2}$.

3. Substituting back gives $S=2\pi r^2+2\pi r(\frac{60}{r^2})=2\pi r^2+\frac{120\pi}{r}$ $=2\pi(r^2+\frac{60}{r})$.

Sol 11

1. The cost of the top and bottom is given by $12(2\pi r^2)$, and the cost of the side is given by $9(2\pi rh)$, so the total cost is given by $C=24\pi r^2+18\pi rh$.

2. The volume of the cylinder is the area of the base multiplied by the height, so

$V=\pi r^2 h=54\pi$ and $h=\frac{54}{r^2}$.

3. Substituting back gives $C=24\pi r^2+18\pi r(\frac{54}{r^2})=24\pi r^2+\frac{972\pi}{r}$ $=12\pi(2r^2+\frac{81}{r})$.

Sol 12

1. We know that $V=\pi r^2 h$ for the cylinder.

2. The total surface area is $80\pi$ square inches, so $S=2\pi r^2+2\pi rh=80\pi$ and therefore

$r^2+rh=40$. Solving for $h$ gives $rh=40-r^2$ and $h=\frac{40-r^2}{r}$.

3. Substituting back gives

$V=\pi r^2(\frac{40-r^2}{r})=\pi r(40-r^2)=\pi(40r-r^3)$.

Sol 13

1. We know that $L=a+b$, where

2. $a^2=x^2+6^2$ by the Pythagorean Theorem, so $a=\sqrt{x^2+36}$.

Using similar triangles, $\frac{b}{4}=\frac{a}{x}$, so $b=\frac{4a}{x}=\frac{4\sqrt{x^2+36}}{x}$.

3. Substituting back gives $L=\sqrt{x^2+36}+\frac{4\sqrt{x^2+36}}{x}$ $=\sqrt{x^2+36}(1+\frac{4}{x})$.

Sol 14

1. We have that $T=T_1+T_2$, where $T_1$ is the time he walks off the road and $T_2$ is the time he walks along the road.

2. Using the formula $D=rt$, we get that $t=D/r$; so

$T_1=\frac{\sqrt{x^2+16}}{3}$ and $T_2=\frac{10-x}{5}$.

3. Substituting back gives $T=\frac{\sqrt{x^2+16}}{3}+\frac{10-x}{5}$.