Loyola University Chicago
Math 263, Fall 2011
Worksheet 3
Problem 1. Find all local minima, local maxima, and saddle points of
f (x, y) = 6x2 − 2x3 + 3y 2 + 6xy
Solution: to find critical points, set ∇f equal to h0, 0i and solve for (x, y). fx = 12x − 6x2 + 6y,
fy = 6y +6x, so fy = 0 gives y = −x, plug this into fx = 0 to get 12x−6x2 −6x = 0, so 6x−6x2 = 0,
so 6x(1 − x) = 0, so x = 0 or x = 1. Because y = −x, get two critical points: (0, 0) and (1, −1).
To test critical points, use the “second derivative” test. fxx = 12 − 12x, fyy = 6, fxy = 6. At
(0, 0), fxx = 12 > 0, fxx fyy − (fxy )2 = 36 > 0, so (0, 0) is a local minimum. At (0, 0), fxx = 0,
fxx fyy − (fxy )2 = −36 < 0, so (1, −1) is a saddle point.
Problem 2. Consider the function f (x, y) from Problem 1 and find its absolute minimum and
absolute maximum on the triangle between x = 2, y = −2, y = x − 1.
Solution: First, it is a very good idea to carefully sketch the triangle.
Second, look for candidates for the absolute minimum and maximum in the interior of the triangle.
Those candidates have to be local minima or local maxima. The only local minimum we found in
Problem 1 is at (0, 0), and (0, 0) is not in the triangle. There are no local maxima at all. Hence, the
absolute minimum and the absolute maximum are not attained in the interior of the triangle. They
are attained on the boundary.
To deal with the boundary, analyze it edge by edge. Consider the bottom edge, where y = −2 and
−1 ≤ x ≤ 2. Then f (x, −2) = 6x2 − 2x3 + 12 − 12x. Look for critical points by setting the derivative
equal to 0: (6x2 − 2x3 + 12 − 12x)0 = 12x − 6x2 − 12 = −6(x2 − 2x + 2) = −6((x − 1)2 + 1) = 0.
There are no solutions to this equation, so the only candidates for an absolute minimum and absolute
maximum along this edge are at the endpoints, x = −1 and x = 2. Check the values: f (−1, −2) = 32,
f (2, −2) = −4.
Consider the right edge, where x = 2 and −2 ≤ y ≤ 1. Then f (2, y) = 3y 2 + 12y + 8. Look
for critical points by setting the derivative equal to 0: (3y 2 + 12y + 8)0 = 6y + 12 = 0, get y = −2.
Compare this to the values at the endpoints y = −2 and y = 1: f (2, −2) = −4, f (2, 1) = 23.
Consider the diagonal edge, where y = x − 1 and −1 ≤ x ≤ 2. Then f (x, x − 1) = 6x2 − 2x3 +
3(x − 1)2 + 6x(x − 1) = −2x3 + 15x2 − 12x + 3. Look for critical points by setting the derivative equal
2
to 0: (−2x3 +15x
−12x+3)√0 = −6x2 +30x−12
= −6(x2 −5x+2) = 0. Solve this quadratic
equation
√
√
√
2
5± 12
5+ 12
5− 12
5± 5 −4 1 2
= 2 . Now, x = 2 > 2, so this can be ignored. x = 2 > 2 ≈ 0.7
to get x =
2
and this is in the interval −1 ≤ x ≤ 2. Compare value f (x, x − 1) at this x, which is approximately
−3, to f (−1, −2), f (2, 1) which we calculated before anyway.
We have the following candidates for the absolute minimum and the absolute maximum (the last
one is approximate):
f (−1, −2) = 32, f (2, −2) = −4, f (2, 1) = 23, f (.7, −.3) ≈ −3
Pick the largest and the smallest value; find the absolute maximum of 32 at (−1, −2), the absolute
minimum of −4 at (2, −2).
Problem 3. Find the maximum value of f (x, y) = 49 − x2 − y 2 on the line x + 3y = 10.
Solution: Somehow, justify to yourself that the maximum value is in fact attained. Then, use
Lagrange multiplier theory to find it. If (x, y) is where a maximum is attained, and if ∇g(x, y) 6= 0,
where g(x, y) = x + 3y − 10, then ∇f (x, y) = λ∇g(x, y) for some real number λ. Thus
h−2x, −2yi = λh1, 3i, −2x = λ, −2y = 3λ, so − 2y = −6x, so y = 3x.
Because x + 3y = 10 and y = 3x, x = 1 and y = 3. Hence, the maximum is attained at (1, 3), and
its value is 39.
Alternative solution: Recall that x2 + y 2 is the distance of the point (x, y) from the origin (0, 0).
Thus, maximum of 49 − x2 − y 2 , vaguely, corresponds to 49 minus minimum of x2 + y 2 . Hence, solve
this problem now: Find the point on the line x + 3y = 10 which is closest to the origin. Do this
however you want, find that (1, 3) is the closest point.
Problem 4. A closed rectangular box is to have volume V cm3 . The cost of the material used in
the box is a cents/cm2 for top and bottom, b cents/cm2 for front and back, and c cents/cm2 for the
remaining sides. What dimensions minimize the total cost of materials?
Solution: let the dimensions of the box be x, y, and z. The total cost of the box is C = axy +
bxz + cyz. The volume V = xyz is constant. We need to solve ∇C = λ∇V :
hay + bz, ax + cz, bx + cyi = λhyz, xz, xyi
with the extra knowledge that xyz = V . Note that ay + bz = λyz, ax + cz = λxz imply that
y(a − λz) = −bz, x(a − λz) = −cz, and dividing these equations one by the other yields y/x = b/c.
Hence, y = bx/c. Similarly, ay + bz = λyz, bx + cy = λxy imply z(b − λy) = −ay, x(b − λy) = −cy,
and so z/x = a/c, so z = ax/c. Then
ab
bx ax
= 2 x3 ,
c c
c
r
r
r
r
2
2
2
b 3 c2
3 c
3 b
3 a
hence x =
V , and then y =
V =
V, x=
V.
ab
c ab
ac
bc
V = xyz = x