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2 years ago

Find y. A. 63–√ B. 33–√ C. 23√3 D. 3

Mathematics

1 answer:

vagabundo [1.1K]2 years ago

5 0

<h3>Answer: D. 3</h3>

This is a 30-60-90 triangle. The shortest leg (opposite the smallest angle) is half that of the hypotenuse, so y = 6/2 = 3.

Extra info: The long leg is sqrt(3) times the short leg, so x = y*sqrt(3) = x = 3*sqrt(3)

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2 years ago

Find a function where f(0)=2 and f(1)=2

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Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$

$f(x) = 2cos(2\pi x)$

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

$f(x)= 2+sin(2\pi x)$

Sky is the limit.

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2 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).