[MOVED] Finite/Discrete Math: express perim. as fcn of width

[langhost]

I'm totally new to Finite math and there is this one question I just can't wrap my head around! It's probably laughably easy to someone who knows what they're doing since I'm in the intro part of the course. I even have the answer--I just don't know how on earth they managed to get that function out of the information given.

Problem: The area of a rectangle is 25 square inches. Express the perimeter P(w) as a function of the width w

Answer: P(w) = 2w + 50/w

(for quick reference Perimeter is 2(length + width) and Area is length * width)

Thanks!

 

[stapel]

Re: [MOVED] Finite/Discrete Math: express perim. as fcn of w

The area of a rectangle is 25 square inches. Express the perimeter P(w) as a function of the width w

Work with the basic formulas and the given information:

. . . . .length "L"
. . . . .width "w"
. . . . .perimeter "P"
. . . . .area "A"

. . . . .A = Lw = 25
. . . . .L = 25/w

. . . . .P = 2L + 2w
. . . . .P = 2[25/w] + 2w

Simplify.

Note: The rectangle in question does not need to be a square. Any rectangle of area 25 square units will do.

Eliz.

 

[pka]

Sorry, I miss read the problem. It is about a rectangle.
So, perimeter is 2W+2L and Area is A=WL or WL=25 therefore L=(25/W).
Thus A(W)=2W+2(25/W)=2W+50/W.

 

[langhost]

Ah! You guys are awesome! It all makes sense now! Thanks!