Question

Under the condition 2x²+y² = 4 for real numbers x,y , the maximum value of 4x + y² is = ? and the minimum one is ??
Maximum = 6
Minimum = -4 * sqrt(2)
I found this results by hit and trial, is there any other methods? Thanks

Answer 1

I can see what you did, by graphing the your answers to the equations, but I don't know why you did it.

I just don't understand the question or what it is asking you to find.

Answer 2

@retirEEd
The condition is 2x²+y² = 4 and x y has to be real numbers. So I did a few math to check the possible x and y values... I would like to know if there is another method to solve this, another way of thinking that I can't see ^^

Answer 3

first one says \[y^2=4-2x^2\]

Answer 4

under those conditions, \[4x+y^2=4x+4-2x^2\] i think

Answer 5

hmm maybe that is not right

Answer 6

well it does give the max of 6, the second coordinate of the vertex. wonder how we find the min?

Answer 7

The constraint: \(\color{black}{g(x,y)=2x^2+y^2= 4}\)
Function to maximize: \(\color{black}{f(x,y)=4x+y^2}\)

\(\color{black}{\nabla f=\lambda \nabla g}\)
\(\color{black}{\langle f_x,~f_y\rangle =\lambda \langle g_x,g_y\rangle}\)
\(\color{black}{\langle 4,~2y\rangle = \langle 4\lambda x,~4\lambda y\rangle}\)

So the system you have is:
\(\color{black}{4=4\lambda x}\)
\(\color{black}{2y=4\lambda y}\)
\(\color{black}{2x^2+y^2= 4 }\)

Answer 8

Well, if you learned Lagrange multipliers (that's the first thing that comes to my mind)

Answer 9

@SolomonZelman hmm... no I didn't, I just found the maximum with Yv = -delta/4a , and manually found a few values for x and y in order to find the minimum.