Mathematics, 29.07.2019 21:00 SpittingFatLama

Find the absolute maximum and minimum values of f on the set d. f(x, y) = xy2 + 2, d = {(x, y) | x ≥ 0, y ≥ 0, x2 + y2 ≤ 3}

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Answer from: Rosa2602

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:

$\mathbf{f(x,y) = xy^2 + 2}$

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

$\mathbf{f_x(x,y) = y^2}$

$\mathbf{f_y(x,y) = 2xy}$

Substitute 0 for y in the first equation.

So, we have:

$\mathbf{x = 0}$

This means that:

$\mathbf{Critical\ Point = f(0,0) = 0}$

Next, we calculate the extreme values.

We have:

$\mathbf{x^2 + y^2 \le 3}$

Express as an equation

$\mathbf{x^2 + y^2 = 3}$

Make y the subject

$\mathbf{y^2 = 3 - x^2}$

Take square roots

$\mathbf{y = \sqrt{3 - x^2}}$

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

So, we have:

$\mathbf{f(x) = x(3 -x^2) + 2}$

Open brackets

$\mathbf{f(x) = 3x -x^3 + 2}$

Differentiate

$\mathbf{f(x) = 3 -3x^2 }$

Set to 0

$\mathbf{3 -3x^2 =0}$

Collect like terms

$\mathbf{3x^2 =3}$

Divide through by 3

$\mathbf{x^2 =1}$

Take square roots of both sides

$\mathbf{x =\±\sqrt1}$

$\mathbf{x =\±1}$

The domain is given as:

$\mathbf{x \ge 1}$

So, the possible value of x is:

$\mathbf{x =1}$

Substitute 1 for x in $\mathbf{y = \sqrt{3 - x^2}}$

$\mathbf{y = \sqrt{3 - 1^2}}$

$\mathbf{y = \sqrt{3 - 1}}$

$\mathbf{y = \sqrt{2}}$

So, we have:

$\mathbf{(x,y) = (1,\sqrt{2})}$

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

Substitute these values in f(x,y)

$\mathbf{(x,y) = (0,0) }$

$\mathbf{f(0,0) = 0 \times 0^2 + 2 = 2}$

$\mathbf{(x,y) = (1,\sqrt{2})}$

$\mathbf{f(1,\sqrt 2) = 1 \times (\sqrt 2)^2 + 2 = 4}$

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

Read more about absolute minimum and maximum at:

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Answer from: alicia234

$\implies\begin{cases}f_x=y^2=0\\f_y=2xy=0\end{cases}$

which occurs whenever y=0

. Under either condition, we have f(x,0)=f(0,y)=2

. The equations above admit no other critical points on the surface, so we consider the boundaries. We already have x=0

and

accounted for, so let's look at the circular part.

We can parameterize this by

$\begin{cases}x=\sqrt3\cos t\\y=\sqrt3\sin t\end{cases}$

with $0\le t\le\dfrac\pi2$ . So now

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

which can be written in terms of cosine alone as

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

Now, we get critical points at

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

We omit t=0

, because that gives us F(0)=2

again. Meanwhile, when $t=\arccos\dfrac1{\sqrt3}$ , we get $F\left(\arccos\dfrac1{\sqrt3}\right)=4$ .

Finally, we compute the corresponding critical point in terms of x,y

. We have

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

So to summarize, minimum values of 2 are obtained anywhere along the borders where x=0

, and a maximum value of 4 at $(x,y)=(1,\sqrt2)$ .

Answer from: Quest

answer: it's upside down and i can't read it

step-by-step explanation:

Me with this . im really bad at these types of problems!

Answer from: Quest

we have three similar triangles, because each has a right angle and shares an angle. let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

$(5 + x)/6 = 6/x$

$x(x+5) = 36$

$x^2 +5x - 36 = 0$

$(x+9)(x-4) = 0$

we reject the negative root and conclude x=4

answer: 4

If bc = 6 and ad = 5, find dc. a) 4 b) 4.5 c) 7.2

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