Area of Rectangle As Function of x

[nycmathguy]

Section 1.4
Question 63

See attachment.

How is this done?

 

[MathLover1]

1) Find the relationship betwenn y and x.

You do this trhough the slope equation.

You have three points. (0,b) , (2,1) , and (a,0) or (since a=x and b=y) your points are (0,y) , (2,1) , and (s,0)

Slope between (0,y) and (2,1) = [y - 1] / [ 0 -2] = [1 - y] / 2

Slope between (2,1) and (x,0) = [1 - 0] / [2 -x] = 1 / [2 - x]

2) Now establish the formula for the area of the triangle:

(1 / 2) base*heigth = (1/2) xy

3) replace y with x / (x - 2)

area = (1/2) x [ x / (x-2)] = [x^2] / [2(x-2)] = x^2 / (2x - 4)

Answer:
area = x^2 / (2x - 4)

Domain:

x > 2 to y be positive,

Answer: x element (2, ∞)

 

[nycmathguy]

1) Find the relationship betwenn y and x.

You do this trhough the slope equation.

You have three points. (0,b) , (2,1) , and (a,0) or (since a=x and b=y) your points are (0,y) , (2,1) , and (s,0)

Slope between (0,y) and (2,1) = [y - 1] / [ 0 -2] = [1 - y] / 2

Slope between (2,1) and (x,0) = [1 - 0] / [2 -x] = 1 / [2 - x]

2) Now establish the formula for the area of the triangle:

(1 / 2) base*heigth = (1/2) xy

3) replace y with x / (x - 2)

area = (1/2) x [ x / (x-2)] = [x^2] / [2(x-2)] = x^2 / (2x - 4)

Answer:
area = x^2 / (2x - 4)

Domain:

x > 2 to y be positive,

Answer: x element (2, ∞)

1. Where did s come from in the point (s, 0)?

2.You said to replace y with x/(x - 2). Where did
x/(x - 2) come from?

3. You said "x > 2 to y be positive." Explain. In words, x is greater than 2 to y be positive." What did you include the variable y in your domain answer?

 

[MathLover1]

1. typo , (s, 0) is actually (x, 0)

2.
from slopes

Slope between (0,y) and (2,1) = [y - 1] / [ 0 -2] = [1 - y] / 2

Slope between (2,1) and (x,0) = [1 - 0] / [2 -x] = 1 / [2 - x]

recall that slopes are equal

[1 - y] / 2=1 / [2 - x] ..........solve for y

1 - y =2 / [2 - x]

y=1 - 2 / [2 - x]

y=[2 - x]/ [2 - x]- 2 / [2 - x]

y=[2 - x- 2] / -[x - 2]


y=- x / -[x - 2]

y=x / [x - 2]

now substitute y in equation for area= (1/2) xy


3.
denominator (2x - 4) will be equal to zero if x=2, so x must be greater than 2
since y=area => area cannot be negative number

 

[nycmathguy]

1. typo , (s, 0) is actually (x, 0)

2.
from slopes

Slope between (0,y) and (2,1) = [y - 1] / [ 0 -2] = [1 - y] / 2

Slope between (2,1) and (x,0) = [1 - 0] / [2 -x] = 1 / [2 - x]

recall that slopes are equal

[1 - y] / 2=1 / [2 - x] ..........solve for y

1 - y =2 / [2 - x]

y=1 - 2 / [2 - x]

y=[2 - x]/ [2 - x]- 2 / [2 - x]

y=[2 - x- 2] / -[x - 2]


y=- x / -[x - 2]

y=x / [x - 2]

now substitute y in equation for area= (1/2) xy


3.
denominator (2x - 4) will be equal to zero if x=2, so x must be greater than 2
since y=area => area cannot be negative number

I got it. Thanks.