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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).

8 0

4 years ago

Tricia starts school at 7am and has lunch at 12pm. She wants to make sure she has something to eat in between. Determine what ti

Leona [35]

The time in the exactly in the middle would be 9:30AM

4 0

4 years ago

Read 2 more answers

Can someone please help me.

Katena32 [7]

Given
- f(n) values for n=1,2,3,4
- possible candidates for the function

Solution:
Method: Evaluate some of the values, for each function. A function with ANY value not matching the given f(n) values will be rejected.

N=1, f(n)=4
f(1)=4-3(1-1)=4
f(1)=4+3^(1+1)=4+3^2=4+9=13 ≠ 4 [rejected]
f(1)=4(3^(n-1))=4(3^0)=4
f(1)=3(4^(n-1))=3(4^0)=3*1=3 [rejected]

N=2, f(n)=12
f(1)=4-3(2-1)=4-3(1)=1 ≠ 12 [rejected]
[rejected]
f(1)=4(3^(2-1)=4*3^1=4*3=12
[rejected]

Will need to check one more to be sure
N=3, f(n)=3
[rejected]
[rejected]
f(3)=4(3^(n-1))=4(3^(3-1))=4(3^2)=4*9=36 [Good]
[rejected]

Solution: f(n)=4(3^(n-1))

6 0

4 years ago

Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from th

fredd [130]

Using continuity concepts, the answers are given by:

The function is right-continuous at x = 0.

The function is left-continuous at x = 1.

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A function f(x) is continuous at x = a if:

$\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)$

If $\lim_{x \rightarrow a^{-}} f(x) = f(a)$ , the function is left-continuous.
If $\lim_{x \rightarrow a^{+}} f(x) = f(a)$ , the function is right-continuous.
For the given function, the continuity is tested at the points in which the definitions are changed, x = 0 and x = 1.