Math 113 - Lagrange Multipliers Worksheet Answers - 10/14/15
1. (3/5, 4/5) is where it is largest. (−3/5, −4/5) is where it is smallest. The largest value is 5,
the smallest is −5.
2. We have L = 240 and K = 120.
3. For points (a, b) on the circle, ∇g(a, b) is perpendicular to the circle. Therefore, for the
equation ∇f = λ∇g to be satisfies, it must be the case that ∇f is perpendicular to the circle
as well. There are two such points, one near the bottom of the circle and one near the top.
You can (more or less) follow ∇f as you move from the point on the bottom of the circle to
the point on the top of the circle where the circle and ∇f are perpendicular. As you do this,
f is increasing (since ∇f points in the direction of greatest increase). Thus, the point on the
bottom is a minimum and the point on the top is a maximum.
4. The idea here is to, say, make g(x1 , x2 , . . . , xn ) = x21 + x22 + · · · + x2n and hold this constant.
Then, you want to minimize f (x1 , x2 , . . . , xn ) = x31 + x32 + · · · + x3n . If at the minimum value,
the inequality is true, then it must be true always!
5. (b) The point closest to the origin is (0, 0).
(c) In essence, the solution in part (a) is parametrizing the curve 4y = x2 by using ~r(y) =
√
h2 y, yi, but this only gives a point on the parabola if y ≥ 0. For this reason, the right thing
to do is minimize h(y) with the additional constraint that y ≥ 0. Since h0 (y) > 0 for all y ≥ 0,
the minimum value occurs when y = 0 and h(y) = 1.
p
√
6. ha = 3 + 3.
1