MATH 53 HW 7 SOLUTIONS 7/14
#14.2 45:
Assume we have a unit vector u = ha, bi so that Du f (0, 2) = 1. We can express this
derivative as a dot product between u and ∇f (0, 2). For general (x, y)
∇f = −y 2 e−xy , e−xy − xy2−xy
√
so ∇f (0, 2) = h−4, 1i. Then Du f (0, 2) = ha, bi · h−4, 1i = 1 and |u| = a2 + b2 = 1 so
we have the system of equations
−4a + b = 1
a2 + b 2 = 1
Solving the first equation for b and plugging into the second we see that a2 + (1 + 4a)2 =
17a2 + 8a + 1 = 1 ⇒ a(17a + 8) = 0 so a = 0 or −8/17. Plugging these into the first
equations to find b we see that u = h0, 1i or h−8/17, −15, 17i. It is easy to check these
by taking their dot product with h−4, 1i.
#14.7 45:
Suppose that the box has dimensions given by x, y and z. Then the volume V = xyz. In
order that the box the box be inscribed in the sphere, the length of any diagonal must
equal the diameter of the sphere, 2r. To find the length of a diagonal, consider the vector
v = hx, y, zipwhich goes from one corner to the opposite one. Theplength of the diagonal is
then |v| = x2 + y 2 + z 2 . The obvious constraint is then 2r = x2 + y 2 + z 2 . However,
r > 0, so we can also take 4r2 = x2 + y 2 + z 2 , which will simplify calculations. From here
we could use Lagrange multipliers, but the problem comes from 14.7, so let us use the
derivative test instead. If we restrict to the region x ≥ 0, y ≥ 0 and z ≥ 0 then we may
maximize F (x, y, z) = V 2 = (xyz)2 . Plugging in the constraint z 2 = 4r2 − x2 − y 2 we get
f (x, y) = x2 y 2 (4r2 − x2 − y 2 ) = 4x2 y 2 r2 − x4 y 2 − x2 y 4 .
The first derivatives are
fx = 8xy 2 r2 − 4x3 y 2 − 2xy 4 = 2xy 2 (4r2 − 2x2 − y 2 ) = 0
and
fy = 8x2 yr2 − 2x4 y − 4x2 y 3 = 2x2 y(4r2 − x2 − 2y 2 ) = 0.
Now we know that the maximal volume is positive, and if x = 0 or y = 0 the volume is
zero, so we ma assume that x 6= 0 and y 6= 0. Thus we can divide x’s and y’s from the
equations to get
2
4r − 2x2 − y 2 = 0
.
4r2 − x2 − 2y 2 = 0
By taking the difference we see that q
x2 = y 2 . If we plug this into the first equation we
see that 4r2 − 3x2 = 0, so x = y =
4
r.
3
Finally plugging this into the constraint we
q 3
3
4
2
2
2
2
2
have z = 4r − 4/3r − 4/3r = 4/3r , so x = y = z and the volume is
r = 38r√3 .
3
1