<span>We want to optimize f(x,y,z)=x^2 y^2 z^2, subject to g(x,y,z) = x^2 + y^2 + z^2 = 289.
Then, ∇f = λ∇g ==> <2xy^2 z^2, 2x^2 yz^2, 2x^2 y^2 z> = λ<2x, 2y, 2z>.
Equating like entries:
xy^2 z^2 = λx
x^2 yz^2 = λy
x^2 y^2 z = λz.
Hence, x^2 y^2 z^2 = λx^2 = λy^2 = λz^2.
(i) If λ = 0, then at least one of x, y, z is 0, and thus f(x,y,z) = 0 <---Minimum
(Note that there are infinitely many such points.)
(f being a perfect square implies that this has to be the minimum.)
(ii) Otherwise, we have x^2 = y^2 = z^2.
Substituting this into g yields 3x^2 = 289 ==> x = ±17/√3.
This yields eight critical points (all signage possibilities)
(x, y, z) = (±17/√3, ±17/√3, ±17/√3), and
f(±17/√3, ±17/√3, ±17/√3) = (289/3)^3 <----Maximum
I hope this helps! </span><span>
</span>
Answer:
The width of the rectangle is 2.3 m
Step-by-step explanation:
Now, when we are asked to find the area of the rectangle, we will use the formula: 
. We should be given with at least two values, and then we can use the above formula and can find the third variable.
Now, we are given that the area of a rectangle is 12.65 
and the length measures 5.5 m
So the width is given as:
So this is the required with of the rectangle.
Answer:
0
Step-by-step explanation:
Apply Only the Outside and Inside Method of the Foil Method.
Add them together
So our b value is 0.
Answer:
Measurement of angle ABC: 120
Measurement of arc ACE: 240
Measurement of arc AB: 60
Step-by-step explanation:
I got it right on my test
