Domain and Range of a function

[EIGHTSIX7]

Homework Statement

If the maximum allowable width, w, of the wall is 100 cm, what are the domain and rand of the function? Remember to include units for both the domain and range.

Homework Equations

w(d)=2d
w(d)=100, so
100=2d, there fore
d=100/2

The Attempt at a Solution

Domain: d (less than or equal to) 50cm
Range: all real numbers




Sorry, I'm a bit rusty on finding D and R. Thanks for the help, here is the original question via PDF, I got part (a). I am just unsure about my D and R ( I am a perfectionist)

ALSO: For part (a) I got : A(d)=((Pi)(d^2))/2, correct?

THANK YOU VERY MUCH.

 

[Accretion]

For part a I got [itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}=d^{2}(1+\frac{\pi}{4})[/itex]

I hope I'm understanding the problem correctly, [itex]d[/itex] is also the diameter of the semi-circle on either side, so [itex]r=\frac{d}{2}[/itex]. I think you are neglecting to add the area of the square inbetween those two semi-circles in your function for part a.

Edit: squaring mistake :/

 

[EIGHTSIX7]

For part a I got [itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}=d^{2}(1+\frac{\pi}{4})[/itex]

I hope I'm understanding the problem correctly, [itex]d[/itex] is also the diameter of the semi-circle on either side, so [itex]r=\frac{d}{2}[/itex]. I think you are neglecting to add the area of the square inbetween those two semi-circles in your function for part a.

Edit: squaring mistake :/

Area of cross section is (Area of Square=s^2)+(2 Semi-circles, which is equivalent to one circle, so Area=(Pi)(r^2))

Area of Square + Area of Circle = Area of Cross section

Since ,d, in diagram is also the sides of the square I have accounted for the area of a square.

So area of a circle in terms of Area of Circle(d)= ((Pi)((d/2)^2))
there fore,

Area of Cross section (d)= (d^2)+((Pi)((d/2)^2))
simplified: A(d)=((Pi)(d^2))/2

 

[Accretion]

Area of cross section is (Area of Square=s^2)+(2 Semi-circles, which is equivalent to one circle, so Area=(Pi)(r^2))

Area of Square + Area of Circle = Area of Cross section

Area of Cross section (d)= (d^2)+((Pi)((d/2)^2))
simplified: A(d)=((Pi)(d^2))/2

I'm not sure I follow your simplification to: [itex]A(d)=\frac{\pi d^{2}}{2}[/itex]

My Reasoning:
Area of square: [itex]d^{2}[/itex]
Area of circle: [itex]\pi r^{2}[/itex]
Radius of circle: [itex]r=\frac{d}{2}[/itex]
[itex]A(d)=d^{2}+\pi r^{2}[/itex]
[itex]A(d)=d^{2}+\pi(\frac{d}{2})^{2}[/itex]
[itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
[itex]A(d)=d^{2}(1+\frac{\pi}{4})[/itex]

 

[EIGHTSIX7]

I'm not sure I follow your simplification to: [itex]A(d)=\frac{\pi d^{2}}{2}[/itex]

My Reasoning:
Area of square: [itex]d^{2}[/itex]
Area of circle: [itex]\pi r^{2}[/itex]
Radius of circle: [itex]r=\frac{d}{2}[/itex]
[itex]A(d)=d^{2}+\pi r^{2}[/itex]
[itex]A(d)=d^{2}+\pi(\frac{d}{2})^{2}[/itex]
[itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
[itex]A(d)=d^{2}(1+\frac{\pi}{4})[/itex]

Which step are you doing from
[itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
to
[itex]A(d)=d^{2}(1+\frac{\pi}{4})[/itex]

 

[Accretion]

Which step are you doing from
[itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
to
[itex]A(d)=d^{2}(1+\frac{\pi}{4})[/itex]

I'm just factoring the [itex]d^{2}[/itex] out of [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex]. Multiply [itex]d^{2}(1+\frac{\pi}{4})[/itex] out using the distributive property and you'll get back to [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex].

 

[EIGHTSIX7]

I'm just factoring the [itex]d^{2}[/itex] out of [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex]. Multiply [itex]d^{2}(1+\frac{\pi}{4})[/itex] out using the distributive property and you'll get back to [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex].

Wow, I'm embarrassed and you are great.
Thanks a lot.