Finding Extrema of a Function with Two Constraints using Lagrange Multipliers - Prof. E. S, Study notes of Mathematics

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Math 114, Chapter 15, Section 8, Question 39

Shuvra Gupta

October 30, 2006

Please refer to the question mentioned above. We see that we are required to find the extremas of a function subject to two constraints, viz. we have to find the extremas of the distance (from the origin) function subject to the fact that the point satisfies the two given equations of the plane and the paraboloid. To spell out things even more explicitly, we are required to find the points of maxima and minima of the function f 0 (x, y, z) =

√

x^2 + y^2 + z^2 subject to the constraints g(x, y, z) = h(x, y, z) = 0, where g(x, y, z) = x + y + 2z − 2 and h(x, y, z) = x^2 + y^2 − z. (Please note that f 0 is as above because it is the distance from the point (x, y, z) to the origin (0, 0 , 0).) Important: We know that we have to consider the gradients of the func- tions above including that of f 0 but as f 0 is the square root of something, its gradient is going to be slightly messy. Now, we note that when√ f 0 (x, y, z) = x^2 + y^2 + z^2 attains its maximum or minimum, the function f (x, y, z) = x^2 + y^2 + z^2 also attains its maximum or minimum respectively. The latter function is in some ways a “simpler” function, and so we will be considering f instead of f 0. (As you see other problems involving finding extrema using Lagrange multi- pliers, you will realise this is in fact what we always do to make things easier- instead of taking the square root of a function, consider the function.)

Now, the method of Lagrange multipliers tells us that the extreme values occur when ∇f is of the form λ∇g + μ∇h for some values of λ and μ. We know,

∇f = 〈 2 x, 2 y, 2 z〉 ∇g = 〈 1 , 1 , 2 〉 ∇h = 〈 2 x, 2 y, − 1 〉

So, ∇f = λ∇g + μ∇h implies 〈 2 x, 2 y, 2 z〉 = 〈λ + 2μx, λ + 2μy, 2 λ − μ〉. Thus, we have the following three equations

2 x = λ + 2μx (1) 2 y = λ + 2μy (2) 2 z = 2 λ − μ (3)

Considering the difference (1)−(2), we get 2(x−y) = 2μ(x−y). This implies that either μ = 1 OR x − y = 0, which is the same as x = y. (It is important here not to ignore any of the conditions.)

Case I: μ = 1. Substituting μ = 1 in equation (1), we get 2x = λ + 2x, implying λ = 0. Substituting the values of λ and μ in equation (3), we get 2z = −1, implying z = −^12. Now, recall that we still haven’t used the two constraints that we had. Using z = −^12 in the constraint h, we get x^2 +y^2 −(−^12 ) = 0, implying x^2 +y^2 = −^12 , which clearly has no solutions as the left hand side is always greater than or equal to 0.

Case II: x = y. We see that we cannot get any additional information from equations (1), (2) and (3). So, we go back to the constraints. Plugging x = y in the constraints and eliminating y, we get

2 x + 2z − 2 = 0 (4) 2 x^2 − z = 0 (5)

Equation (4) simplifies to x+z−1 = 0, implying z = −x+1. Substituting this in equation (5), we get 2x^2 +x−1 = 0. Factorising, we get (2x−1)(x+1) = 0 giving us x = 12 , −1. If x = 12 , y = x = 12 and z = 2x^2 = 12. If x = −1, y = x = −1 and z = 2x^2 = 2. Thus the points are (^12 , 12 , 12 ) and (− 1 , − 1 , 2) and their distances from the

origin are

√ 3 2 and^

√

6 respectively. Thus, the nearest and farthest points (from the origin) are (^12 , 12 , 12 ) and (− 1 , − 1 , 2) rspectively.