Section 13.1Extreme values 4. Refer to Example5 in page746.
5. Refer to Example5 in page746.
7. Refer to Example5 in page746.
11. Refer to Example5 in page746.
14.Refer to Example5 in page746.
19.
f (x, y) = xye−x2+y4, f1(x, y) = y(1− 2x2)e−x2+y4, f2(x, y) = x(1− 4y4)e−x2+y4 For critical points y(1− 2x2) = 0 and x(1− 4y4) = 0. The critical points are
(0, 0), (± 1
√2, 1
√2), (± 1
√2,− 1
√2)
20.Refer to Example5 in page746.
25.Let(x, y, z) be the coordinates of the corner of the box that is in the first octant of space.Thus x, y, z ≥ 0,and xa22 +yb22 +zc22 = 1. The volume of the box is V = (2x)(2y)(2z) = 8cxy
√
1−xa22 −yb22 for x ≥ 0, y ≥ 0 and xa22 +yb22 ≤ 1. For analysis it is easier to deal with V2than with V : V2= 64c2(x2y2−x4ay22)−x2b2y4
27.Differeniate the given equation e(2zx−x2)− 3e(2zy+y2) = 2 with respect to x and y, regarding z as a function of x and y.
28.we will use the second derivative test to classify the two critical points calculated in Exercise 27. To calculate the second partials
A = ∂2z
∂x2, B = ∂2z
∂x∂y, C = ∂2z
∂y2 we obtain
e2zx−x2[(2x∂z
∂x+ 2z−2x)2+ 4∂z
∂x+ 2x(∂)2z
∂x2 −2]−3e2yz+y2[(2y∂z
∂x)2+ 2y(∂)2z
∂x2 ] = 0.
On the other hand , e2zx−x2[(2x∂z
∂y)2+ 2x∂2z
∂y2]− 3e2zy+y2[(2y∂z
∂y + 2z + 2y)2+ 4∂z
∂y + 2y∂2z
∂y2 + 2] = 0.
Finally, e2zx−x2[(2x∂z
∂x+2z−2x)(2x∂z
∂y)+2x ∂2z
∂x∂y+2∂z
∂y]−3e2zy+y2[(2y∂z
∂y+2z+2y)(2y∂z
∂x)+2∂z
∂x+2y ∂2z
∂x∂y] = 0.
1
