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2 years ago

Question 12(Multiple Choice Worth 1 points)

Mathematics

1 answer:

rusak2 [61]2 years ago

7 0

Answer:

$\frac1{25}$

Step-by-step explanation:

Hello!

Let's use a couple of exponent rules:

$\frac{a^x}{a^b} = a^{x - b}$
$a^{-x} = \frac{1}{a^x}$

Let's utilize the first rule and simplify the fraction.

<h3>Simplify</h3>

$\frac{5^4}{5^6}$
$5^{4 - 6}$
$5^{-2}$

Now, we can utilize the second exponent rule, as we have a base to the power of a negative number.

<h3>Simplify</h3>

$5^{-2}$
$\frac1{5^2}$
$\frac1{25}$

The answer is option 1, $\frac{1}{25}$ .

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i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

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If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
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Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

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

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.