Find the measure of a regular 45 agon
2 answers:
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Answer:
The maximum value= 36
Minimum value = - 36
Step-by-step explanation:
Given that
f(x, y, z) = 8 x + 8 y + 4 z
h(x,y,z)=4 x² + 4 y² + 4 z² - 36
From Lagrange multipliers
Δf = λ Δh
Δf = < 8 ,8 , 4>
Δh = < 8 x ,8 y , 8 z>
Δf = λ Δh
So
< 8 ,8 , 4> = < 8 λ x ,8 λ y , 8 λ z>
8 = 8 λ x -------------1
8 = 8 λ y ---- ------2
4 = 8 λ z ----------------3
From equation 1 ,2 and 3
Now by putting the value of x,y and z in the following equation
4 x² + 4 y² + 4 z² = 36
So the value of λ is
When λ = 1/2
x = 1 / λ , y=1 / λ , z= 1 /2 λ
x= 2 , y = 2 , z=1
So
f(x, y, z) = 8 x + 8 y + 4 z
f(2, 2, 1) = 8 x 2 + 8 x 2 + 4 x 1
f(2, 2, 1) =36
When λ = - 1/2
x = 1 / λ , y=1 / λ , z= 1 /2 λ
x= - 2 , y = - 2 , z= - 1
So
f(x, y, z) = 8 x + 8 y + 4 z
f(-2, -2, -1) = 8 x (-2) + 8 x (-2) + 4 x (-1)
f(-2, -2, -1) = - 36
The maximum value= 36
Minimum value = - 36