This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z) = 4x + 4y + 2z; 2x2 + 2y2 + 2z2 = 18

We have ∇f(x,y,z) = ⟨4x3,4y3,4z3⟩ and ∇g(x,y,z) = ⟨2x,2y,2z⟩, so LaGrange’s method gives requires that we solve the following system of equations:

x2 + y2 + z2 = 1
We split into four cases, depending on whether x and y are zero or not:

4x3 = 2λx 4y3 = 2λy 4z3 = 2λz

(1) (2) (3) (4)

(a) x and y are both nonzero. Then equations (1) and (2) tell us that x2 = y2 = λ/2, and putting √√ √√√

this into equations (3) and (4) gives solutions (± 2/2, ± 2/2, 0) and (± 3/3, ± 3/3, ± 3/3). (b) x̸=0buty=0. Thenwehavex2 =λ/2,from(1)andputtingthisinto(4)givesλ/2+z2 =1,

√√ which using (3) gives solutions (±1, 0, 0) and (± 2/2, 0, ± 2/2).

(c) y ̸= 0 but x = 0. This is just like case (b) but with x and y reversed: the solutions are (0, ±1, 0) √√

and (0, ± 2/2, ± 2/2).
(d) x = y = 0. Then equation (4) tells us that z = ±1, so we get the two solutions (0, 0, ±1).

Now we determine which of these points are maxima and minima by simply evaluating f at all these points. We find that the maximum value of f is 1 and occurs at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1),

√√√√√√

while the minimal value is 1/3, and occurs at (± 3/3, ± 3/3, ± 3/3), (± 3/3, ± 3/3, ± 3/3), √√√

and (± 3/3, ± 3/3, ± 3/3). 

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step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

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