Given $f(x,y)=2x^4-xy^2+2y^2,0\le x\le 4, 0\le y\le2$. Find absolute extrema of $f(x,y)$.

You are watching: Find the absolute maxima and minima of the function on the given domain.

I have uncovered $\partial f/\partial x=8x^3-y^2, \partial f/\partial y=-2xy+4y$ and after fixing the equation through letting $\partial f/\partial x=0$ and $\partial f/\partial y=0$, the an essential point room $(0,0), (2,-8) and also (2,-8)$. I"m lost just how to discover the absolute extrema, isn"t any kind of alternative means to resolve this question?

You deserve to follow a recipe because that these questions.

$\quad(1)$: If the preferably or minimum lies ~ above the internal of the domain, climate it need to be a an essential point (that is, its gradient must vanish).

$\quad\quad(1.1)$: To recognize whether a crucial point is a neighborhood maximum or a regional minimum (or saddle), friend may directly compute the values and also compare them, or else employ a greater order test (e.g. Hessian).

$\quad(2)$: If the best or minimum lies on the boundary of the domain, you might use Lagrange multipliers to uncover them.

$\quad\quad(2.1)$: as soon as the boundary of the domain is $1$-dimensional, you might parametrize and also min-max follow me the parametrization; this to reduce the step to a single-variable calculus exercise.

In general, us don"t know a priori even if it is $(1)$ or $(2)$ applies, therefore we need to examine for extrema both in the interior and in the border of the domain.

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answered might 17 "18 in ~ 20:08

FimpellizieriFimpellizieri
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Recall that we additionally need come look for the extrema on the boundary, in this case since we have actually not an important points in the interior of the domain the extrema points are on the boundary, then we have to check

$f(0,y)$ and also $f(4,y)$ because that $0\le y\le2$

$f(x,0)$ and also $f(x,2)$ for $0\le x\le4$

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answered might 17 "18 in ~ 20:07

useruser
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Once you"ve found the extrema, plug them into $f$ and see which one returns the shortest value.

You"ll also need to make sure $f$ isn"t reduced on the border of her domain.

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answered may 17 "18 at 20:07

NicNic8NicNic8
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Since your set $E=\(x,y)\in\princetoneclub.orgbbR^2:0\leq x\leq4,0\leq y\leq2\$ is compact and also $f$ is consistent on $E$ (being a polynomial), we recognize that $f$ suspect a an international minimum and also a global maximum top top $E$.

You can uncover the extrema in the following way.

First friend look for extrema in the interior by calculating the first partial derivatives. Setup both same to zero and solving yields critical points.

This must be sufficient to resolve this exercise.

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answered might 17 "18 in ~ 20:12

Václav MordvinovVáclav Mordvinov
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You deserve to avoid the constraints making

$$u = 4\left(\frac\sin(\phi)+12\right)\\v = 2\left(\frac\sin(\eta)+12\right)$$

with this adjust of variables the trouble reads

$$\min\max f(\phi,\eta) = 2u^4-uv^2+2v^2 = 2u^4-(u-2)v^2$$

The stationary problems give

$$\nabla f = (f_\phi,f_\eta) = \left\{\beginarrayrcl128 (\sin (\phi )+1)^3 \cos (\phi )-2 (\sin (\eta )+1)^4 \cos (\phi )&=&0\\-8 (\sin (\eta )+1)^3 \cos (\eta ) (\sin (\phi )+1)-4 (\sin (\eta )+1) \cos (\eta )& = & 0\endarray\right.$$

or

$$\left\{\beginarrayrcl\cos(\phi) & = & 0\\\sin(\eta) & = & 0\\64 (\sin (\phi )+1)^3-(\sin (\eta )+1)^4&=&0\\2 (\sin (\eta )+1)^2 (\sin (\phi )+1)+1& = & 0\endarray\right.$$

Calling

$$p = \sin(\phi)+1\\q=\sin(\eta)+1$$

the mechanism last two equations read

$$64p^3-q=0\\2p q^2+1=0$$

and hence the values for $\phi, \eta$ are conveniently obtained

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answered might 17 "18 in ~ 21:20
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