Help with precalc word problem

[VP1]

Hi all, I'm taking my first semester of calculus and am in the process of working on a precalc review homework problem. I think I am doing it right but would like some confirmation and also a little help with one of the questions.
The question is this-

A piece of wire 24 inches long is bent into the shape of a rectangle having width x and length y.

a. Express y as a function of x.

I come up with 2y + 2x = 24 which ends up as y = -x + 12.

b. Express the area A of the rectangle as a function of x.

So then A= x*y and since y= -x+12 we have A= x(-x+12) an multiplying gives us A= -x² + 12x. Does that look ok?

c. Show that the area A is the greatest if the rectangle is a square.

This is is causing me some trouble. I'm not really sure what to do with the above info at this point.. I've thought about locating the vertex which give me x=6 but I don't know if that is of any help to my problem.

I'd appreciate any advice!
Thanks

 

[Deleted member 4993]

Hi all, I'm taking my first semester of calculus and am in the process of working on a precalc review homework problem. I think I am doing it right but would like some confirmation and also a little help with one of the questions.
The question is this-

A piece of wire 24 inches long is bent into the shape of a rectangle having width x and length y.

a. Express y as a function of x.

I come up with 2y + 2x = 24 which ends up as y = -x + 12. Correct

b. Express the area A of the rectangle as a function of x.

So then A= x*y and since y= -x+12 we have A= x(-x+12) an multiplying gives us A= -x² + 12x. Does that look ok? Correct

c. Show that the area A is the greatest if the rectangle is a square.

This is is causing me some trouble. I'm not really sure what to do with the above info at this point.. I've thought about locating the vertex which give me x=6 but I don't know if that is of any help to my problem.

I'd appreciate any advice!
Thanks

A= -x² + 12x → y = -x² + 12x this is an equation of a parabola - where is the vertex of this parabola? The "x" value at that vertex will give you maximum "y" (= A).

 

[VP1]

Thanks for the reply. So I think I was on the right track after thinking about your reply. Locating the vertex gives us x=6. F(6) is -36. This is an upwards opening parabola and thus has no max value. Or so I think .. But how does that necessairly show that it is the best shape?

Thanks again!

 

[wjm11]

A= -x² + 12x → y = -x² + 12x this is an equation of a parabola - where is the vertex of this parabola? The "x" value at that vertex will give you maximum "y" (= A).

Locating the vertex gives us x=6. F(6) is -36. This is an upwards opening parabola and thus has no max value. Or so I think ...

No, this is a downward facing parabola, and f(6) = 36 (the maximum area)

 

[VP1]

thanks for the clarification. I was trying to divide the -1 out of the expression which was messing up the orientation of the parabola. Got it figured out now