Solutions to exam 2 practice problems

Then evaluate the discriminant at the point (3, 3),
D = ( 108
27 )2 − 4 = 16 − 4 = 12 > 0
Since D > 0 and f ww > 0, the critical point is a local minimum.
The possible values of w, h are w, h > 0. Notice that f(3, 3) = 54.
To check that it is a global minimum, check that in each of the following situations
f(w, h) > 54.
(a) w ≤ 1/4
(b) h ≤ 1/4
(c) w ≥ 1/4, h ≥ 400
(d) h ≥ 1/4, w ≥ 400
Since the region R where 1/4 ≤ w ≤ 400 and 1/4 ≤ h ≤ 400 is closed and bounded f
achieves a global minimum on R. Since there is only one critical point and the value
at that critical point is less than the values of f along the entire boundary the critical
point is a global minimum.
11. Let f(x, y) = x 2 + 6y 2 and let g(x, y) = x + 3y.
(a) Use Lagrange multipliers to find the minimum of f(x, y) subject to the constraint
g(x, y) = 10. Assume that there is a global minimum.
(b) Explain why f(x, y) has no maximum subject to the constraint g(x, y) = 10.
(c) Explain why g(x, y) has a maximum and a minimum subject to the constraint
f(x, y) = 9.
(d) Find the maximum and minimum of g(x, y) subject to the constraint f(x, y) = 9.
(a) Notice I added ”Assume that there is a global minimum” to the question. If we assume
this, then we know it must occur at a point where either the gradients of f and g are
parallel, an endpoint of the constraint or when the gradient of g is zero.
First compute the gradients of f and g :
∇f(x, y) = 2x⃗i + 12y⃗j
∇g(x, y)
= ⃗i + 3⃗j
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