NepanitaNesg3a

2022-05-02

I maximized the function

$f(x,y)=(x+2)(y+1)$

subject to

$4x+6y=130,\phantom{\rule{1em}{0ex}}x,y\ge 0$

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.

$f(x,y)=(x+2)(y+1)$

subject to

$4x+6y=130,\phantom{\rule{1em}{0ex}}x,y\ge 0$

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.

Zain Mccarty

Beginner2022-05-03Added 15 answers

Note that the function $f$ as viewed along the curve $4x+6y=130$ only has one degree of freedom (thus only needs one parameter to be expressed). We can write the curve as $y=\frac{130-4x}{6}$ and use this as a substitution in $f$ so that

$f(x)=(x+2){\textstyle (}\frac{130-4x}{6}+1{\textstyle )}$

along the curve (yes, specifically along the curve, this is important). We can then look at the second derivative ${f}^{\u2033}(x).$ Notice that only the quadratic term of $f(x)$ will contribute to this second derivative so it must be that

${f}^{\u2033}(x)=\frac{{d}^{2}}{d{x}^{2}}{\textstyle [}\frac{-4{x}^{2}}{6}{\textstyle ]}=\frac{-4}{\phantom{\rule{thickmathspace}{0ex}}3}<0\phantom{\rule{thickmathspace}{0ex}}.$

This is sufficient to say that $f$ is concave along the curve $4x+6y=130$ 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 $4x+6y=130$ whereas the Hessian matrix gives us an idea about the curvature of f as a whole surface.

$f(x)=(x+2){\textstyle (}\frac{130-4x}{6}+1{\textstyle )}$

along the curve (yes, specifically along the curve, this is important). We can then look at the second derivative ${f}^{\u2033}(x).$ Notice that only the quadratic term of $f(x)$ will contribute to this second derivative so it must be that

${f}^{\u2033}(x)=\frac{{d}^{2}}{d{x}^{2}}{\textstyle [}\frac{-4{x}^{2}}{6}{\textstyle ]}=\frac{-4}{\phantom{\rule{thickmathspace}{0ex}}3}<0\phantom{\rule{thickmathspace}{0ex}}.$

This is sufficient to say that $f$ is concave along the curve $4x+6y=130$ 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 $4x+6y=130$ whereas the Hessian matrix gives us an idea about the curvature of f as a whole surface.

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