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3 years ago

6

Which one correct pls help and thanks

1 answer:

lakkis [162]3 years ago

4 0

<h2>Answer:</h2>

the answer is C. (i think)

:) hope i helped

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4 years ago

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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

pashok25 [27]

Answer with Step-by-step explanation:

We are given that

$f(x,y,z)=10x+10y+3z$

$g(x,y,z)=5x^2+5y^2+3z^2=43$

We have to find the extreme values of the function.

$f_x(x,y,z)=10,f_y(x,y,z)=10,f_z(x,y,z)=3$

$g_x(x,y,z)=10x,g_y(x,y,z)=10y,g_z(x,y,z)=6z$

$\Delta f(x,y,z)=\lambda \Delta g(x,y,z)$

Therefore,

$10=\lambda 10x\implies x=\frac{1}{\lambda}$

$10=10y\lambda\implies y=\frac{1}{\lambda}$

$3=6z\lambda$

$z=\frac{1}{2\lambda}$

Substitute the values in g(x)

Then, we get

$\frac{5}{\lambda^2}+\frac{5}{\lambda^2}+\frac{3}{4\lambda^2}=43$

$\frac{20+20+3}{4\lambda^2}=43$

$\frac{43}{4\lambda^2}=43$

$\lambda^2=\frac{43}{4\times 43}$

$\lambda=\pm\frac{1}{2}$

When $\lambda=\frac{1}{2}$ then,

$x=2,y=2,z=1$

When $\lambda=-\frac{1}{2}$

Then, $x=-2,y=-2,z=-1$

$f(2,2,1)=10+10+3=23$

$f(-2,-2,-1)=-10-10-3=-23$

Hence, maximum value of f(x,y,z) is 23 at (2,2,1) and minimum value of f(x,y,z) is -23 at (-2,-2 ,-1).

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4 years ago

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4 years ago

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