Anybody good with word problems?
Shoot them at me
To be honest I have no Idea on this one... but I think I know someone who can help you. @eric_d @PaulaLovesSchool13 @igreen can you help her?
thank you
Do you know calculus?
Barely Im in Pre-Calculus right now
Damn, no calc then. I know how to do this with derivatives, but not without, sorry.
No I have to solve this with cal because its part of the practice problems I want to understand how to do this.
Well, you start by making a function for it, like it says. Do you know how to start?
I have no idea where to start
Okay, why don't we say that the length of the rectangle is y and the width is x?
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So then you get two equations: A= xy and 2x + y = 600
Do you know how to get a quadratic equation from those two equations?
is it like when you do f(g(x))?
It's just a bit of simple rearranging. Take 2x + y = 600 and isolate for y, then plug that value into the second equation.
y=-2x+600 y=x(-2x+600) like this?
Er, no. More like: 2x + y = 600 y = 600 - 2x (2) Insert (2) into A= xy A = xy A = x(600-2x) A = 600x - 2x^2
You might've just made a typo and written y instead of A.
Because otherwise it's right.
No I messed up before and as I was looking at it I realized where I went wrong
Anyways, so know we have the equation A = 600x - 2x^2. Can you take the derivative of this?
what is derivative?
If you can't, you can always just use a graphing calculator to find the max point.
The max point will be the x-value that gives you the max area.
Then you plug in the x-value into the previous equation to find the max area.
I dont really understand what your talking about.
No wait, I've got, the way you do this without derivatives.
Can you convert the equation into the form y = a(x-h)2 + k
Im lost lol
ohh okay
The k value is the max area.
You should get 45 000
yeah I just got that
Thank you very much
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