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LammettHash
[tex]f(x,y)=xy^2+2[/tex]
[tex]\implies\begin{cases}f_x=y^2=0\\f_y=2xy=0\end{cases}[/tex]

which occurs whenever [tex]y=0[/tex] or [tex]x=0[/tex]. Under either condition, we have [tex]f(x,0)=f(0,y)=2[/tex]. The equations above admit no other critical points on the surface, so we consider the boundaries. We already have [tex]x=0[/tex] and [tex]y=0[/tex] accounted for, so let's look at the circular part.

We can parameterize this by

[tex]\begin{cases}x=\sqrt3\cos t\\y=\sqrt3\sin t\end{cases}[/tex]

with [tex]0\le t\le\dfrac\pi2[/tex]. So now

[tex]f(x,y)=f(x(t),y(t))=F(t)=(\sqrt3\cos t)(\sqrt3\sin t)^2+2[/tex]
[tex]\implies F(t)=3^{3/2}\cos t\sin^2t+2[/tex]

which can be written in terms of cosine alone as

[tex]\implies F(t)=3^{3/2}(\cos t-\cos^3t)+2[/tex]

Now, we get critical points at

[tex]F'(t)=3^{3/2}\sin t(3\cos^2t-1)=0\implies t=0\text{ or }\arccos\dfrac1{\sqrt3}[/tex]

We omit [tex]t=0[/tex], because that gives us [tex]F(0)=2[/tex] again. Meanwhile, when [tex]t=\arccos\dfrac1{\sqrt3}[/tex], we get [tex]F\left(\arccos\dfrac1{\sqrt3}\right)=4[/tex].

Finally, we compute the corresponding critical point in terms of [tex]x,y[/tex]. We have

[tex]\begin{cases}x=\sqrt3\cos\left(\arccos\dfrac1{\sqrt3}\right)=1\\\\y=\sqrt3\sin\left(\arccos\dfrac1{\sqrt3}\right)=\sqrt2\end{cases}[/tex]

So to summarize, minimum values of 2 are obtained anywhere along the borders where [tex]x=0[/tex] or [tex]y=0[/tex], and a maximum value of 4 at [tex](x,y)=(1,\sqrt2)[/tex].
MrRoyal

The absolute minimum and maximum values of a function are the least and highest values, the function can take.

The absolute maximum and minimum values of f are 4 and 2, respectively.

The function is given as:

[tex]\mathbf{f(x,y) = xy^2 + 2}[/tex]

First, we determine the partial derivatives of f(x,y)

[tex]\mathbf{f_x(x,y) = y^2}[/tex]

[tex]\mathbf{f_y(x,y) = 2xy}[/tex]

Substitute 0 for y in the first equation.

So, we have:

[tex]\mathbf{x = 0}[/tex]

This means that:

[tex]\mathbf{Critical\ Point = f(0,0) = 0}[/tex]

Next, we calculate the extreme values.

We have:

[tex]\mathbf{x^2 + y^2 \le 3}[/tex]

Express as an equation

[tex]\mathbf{x^2 + y^2 = 3}[/tex]

Make y the subject

[tex]\mathbf{y^2 = 3 - x^2}[/tex]

Take square roots

[tex]\mathbf{y = \sqrt{3 - x^2}}[/tex]

Substitute [tex]\mathbf{y = \sqrt{3 - x^2}}[/tex] in [tex]\mathbf{f(x,y) = xy^2 + 2}[/tex]

So, we have:

[tex]\mathbf{f(x) = x(3 -x^2) + 2}[/tex]

Open brackets

[tex]\mathbf{f(x) = 3x -x^3 + 2}[/tex]

Differentiate

[tex]\mathbf{f(x) = 3 -3x^2 }[/tex]

Set to 0

[tex]\mathbf{3 -3x^2 =0}[/tex]

Collect like terms

[tex]\mathbf{3x^2 =3}[/tex]

Divide through by 3

[tex]\mathbf{x^2 =1}[/tex]

Take square roots of both sides

[tex]\mathbf{x =\±\sqrt1}[/tex]

[tex]\mathbf{x =\±1}[/tex]

The domain is given as:

[tex]\mathbf{x \ge 1}[/tex]

So, the possible value of x is:

[tex]\mathbf{x =1}[/tex]

Substitute 1 for x in [tex]\mathbf{y = \sqrt{3 - x^2}}[/tex]

[tex]\mathbf{y = \sqrt{3 - 1^2}}[/tex]

[tex]\mathbf{y = \sqrt{3 - 1}}[/tex]

[tex]\mathbf{y = \sqrt{2}}[/tex]

So, we have:

[tex]\mathbf{(x,y) = (1,\sqrt{2})}[/tex]

This means that, the absolute maximum is at: [tex]\mathbf{(x,y) = (1,\sqrt{2})}[/tex], while the absolute minimum is at [tex]\mathbf{(x,y) = (0,0) }[/tex]

Substitute these values in f(x,y)

[tex]\mathbf{(x,y) = (0,0) }[/tex]

[tex]\mathbf{f(0,0) = 0 \times 0^2 + 2 = 2}[/tex]

[tex]\mathbf{(x,y) = (1,\sqrt{2})}[/tex]

[tex]\mathbf{f(1,\sqrt 2) = 1 \times (\sqrt 2)^2 + 2 = 4}[/tex]

Hence, the absolute maximum and minimum values of f are 4 and 2, respectively.

Read more about absolute minimum and maximum at:

https://brainly.com/question/13774780

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