<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>
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Answer:
.58c
Step-by-step explanation:
Find 42% of c
.42c
Then subtract that from c
c - .42c
.58c
.58c is 42% less than c
1 oz=<span>0.12503671065 cup or about 0.13 cups
2 cups is 1 pint
1 oz=0.13/2=0.065
so 128 oz=
128 times 0.065=8.32 pints
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Answer:
x = 18
Step-by-step explanation:
b squared + x squared = 30 squared
The answer would be D
-1 3/7=-10/7
-3 2/3=-11/3
-10/7 x -11/3=110/21
110/21=5 5/21
A negative times a negative =a positive
