I maximized the function f <mo stretchy="false">( x , y <mo stretchy="false">)

NepanitaNesg3a

NepanitaNesg3a

Answered question

2022-05-02

I maximized the function
f ( x , y ) = ( x + 2 ) ( y + 1 )
subject to
4 x + 6 y = 130 , x , y 0
and I found ( x , y ) = ( 16 , 11 ). But the question I'm doing asks me to analyse the "sufficient conditions of second order".

So, I thought I should use some theorem that tells me (16,11) is the max value of f. I don't know if I could use Weierstrass' Theorem here and I believe the hessian matrix doesn't help me either.

Answer & Explanation

Zain Mccarty

Zain Mccarty

Beginner2022-05-03Added 15 answers

Note that the function f as viewed along the curve 4 x + 6 y = 130 only has one degree of freedom (thus only needs one parameter to be expressed). We can write the curve as y = 130 4 x 6 and use this as a substitution in f so that
f ( x ) = ( x + 2 ) ( 130 4 x 6 + 1 )
along the curve (yes, specifically along the curve, this is important). We can then look at the second derivative f ( x ) . Notice that only the quadratic term of f ( x ) will contribute to this second derivative so it must be that
f ( x ) = d 2 d x 2 [ 4 x 2 6 ] = 4 3 < 0 .
This is sufficient to say that f is concave along the curve 4 x + 6 y = 130 and thus an extreme point will achieve a maximum.

The distinction here is that we have just computed a sort of "curvature" of f along 4 x + 6 y = 130 whereas the Hessian matrix gives us an idea about the curvature of f as a whole surface.

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