If 2x+y=10 and A=xy, find the maximum value of A.
From the first equation, we have \(y=10-2x\). Then, \(A(x)=x(10-2x)=10x-2x^2\). Now, I assume you know how to find the maximum of this quadratic. :)
I did it this way but the answer isn't correct! :(
The maximum value of A is attained when x=5/2. There, A=25-(25/2)=25/2. You mean to say this isn't correct?
Your answer is correct. How did you do it? I didnt get it. I did it this way: dA/dx= 10 - 4x Then a second derivative for the the max value : dA ^2/ dx^2 =-4
And i got a wrong answer.
\(\Large \frac{d^2A}{dx^2}=-4\) just implies that whatever critical point you get will yield a maximum. (Second Derivative Test) Now, for said critical point, you need to set \(\Large \frac{dA}{dx}=0\), then chug whatever x you get here into A(x).
Oook !! I got it! Thanks alot and sorry for asking again and again! :)
It's okay. :)
:)
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