Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (angelwings996):

Rectangle R has varying length l and width w but a constant perimeter of 4 ft. A. Express the area A as a function of l. What do you know about this function? B. For what values of l and w will the area of R be greatest? Give an algebraic argument. Give a geometric arguement.

OpenStudy (anonymous):

Well, we know that: \[ 2l+2w=P = 4 \]Because they give us a perimeter constraint. This allows us to find \(w\) as a function of \(l\). Can you start by doing that?

OpenStudy (anonymous):

You need to solve for \(w\).

OpenStudy (angelwings996):

am I just solving for w from this equation?

OpenStudy (anonymous):

For now, you are.

OpenStudy (angelwings996):

Okay hold on one second

OpenStudy (angelwings996):

Wait, how do you solve with two equal signs?

OpenStudy (anonymous):

Only use one of them... in this case only use: \[ 2w+2l=4 \]

OpenStudy (angelwings996):

Ok

OpenStudy (angelwings996):

Is it w=2-l ?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Now since you know that: \[ A = w\times l \]and \(w=2-l\) You can get \(A\) in terms of only \(l\).

OpenStudy (anonymous):

This will finish part (a)

OpenStudy (anonymous):

So what is \(A\) as a function of \(l\).

OpenStudy (angelwings996):

Do I substitute 2-l for w in A=wl ?

OpenStudy (anonymous):

Yes

OpenStudy (angelwings996):

then I got \[A=2-l ^{2}\]

OpenStudy (anonymous):

When you substitute it in, it should have parenthesis around it.

OpenStudy (anonymous):

Meaning that you'd have to use the distributive property to simplify it.

OpenStudy (anonymous):

\[ A = (2-l)l \]

OpenStudy (angelwings996):

Wait, I thought I would solve it and it would become 2-l^2

OpenStudy (anonymous):

Yeah, but your algebra is wrong.

OpenStudy (angelwings996):

Oh, nevermind...I see what I did wrong

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!