Use lagrange multipliers to find the maximum and the minimum values of the function f(x, y,z)=yz+xy, subject the constraints xy=1 and y2+z2 =1

You can use Lagrange multipliers, but you don't need to in order to find the function's extremes. The first constraint tells you

... f(x, y, z) = yz + 1

The second constraint tells you

... f(x, y, z) = y·(±√(1-y²)) +1

A graph of this function shows it to have

... a minimum of 0.5

... a maximum of 1.5.

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The Lagrangian can be written as

... L = xy +yz +α(xy -1) +β(y² +z² -1)

Differentiating and setting partial derivatives to zero gives the 5 equations

... ∂L/∂x = 0 = y +αy

... ∂L/∂y = 0 = x + z +αx +2βy

... ∂L/∂z = 0 = y +2βz

... ∂L/∂α = 0 = xy -1

... ∂L/∂β = 0 = y² +z² -1

The first equation tells you α = -1.

Substituting that into the second equation gives one that is symmetrical with the third equation:

... z + 2βy = 0

... y + 2βz = 0

Adding these two equations gives

... (y+z)(1 +2β) = 0

which has solutions

... z = -y ... and ... β = -1/2

Using the latter in either the second or third equation gives

... z = y

Plugging these into the second constraint (the 5th equation), you get

... y² = z² = 1/2

Since z = ±y, yz = ±y² = ±1/2 and the extremes of the objective function become

... f(x, y, z) = ±1/2 + 1 = {1/2, 3/2}

That is ...

... f(-√2, -1/√2, -1/√2) = 3/2 . . . maximum

... f(-√2, -1/√2, 1/√2 = 1/2 . . . . . minimum

... f(√2, 1/√2, 1/√2) = 3/2 . . . . . maximum

... f(√2, 1/√2, -1/√2) = 1/2 . . . . . minimum

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.