Help with calculus homework?

(1)

xy = 4

y = 4/x

Let z = x² + y³

z = x² + (4/x)³

z = x² + 64/x³

dz/dx = 2x - 192/x^4

Let dz/dx = 0.

2x - 192/x^4 = 0

2x^5 - 192 = 0

x = 96^(1/5) ≈ 2.49

This is a local minimum. However, it is not a global minimum.

The function z is continuous on (-∞, 0).

As x → -∞, z → ∞.

As x → 0- , z → -∞.

The range of z is all real numbers, including all positive numbers. There is no smallest positive number, so the function does not have a minimum positive value.

(2)

u + v = 6

v = 6 - u

u²v = u²(6 - u) = 6u² - u³

This is a polynomial of odd degree, so its range is all real numbers. It has no maximum.

1. If xy=4,then when

x*y can only be:

1*4; so it's (x^2)+(y^3) = (1^2)+(4^3) = 1+64=65

2*2 (2^2)+(2^3)=4+8=12

4*1 (4^2)+(1^3)=17

so minimum positive value of (x^2)+(y^3), if xy=4 is 12.

2. Using substitution again, (u^2)v

u+v=6

u v can only be

0 6 = 0 in formula

1 5 = 1*5=5

2 4 = 4*4=16

3 3 = 9*3 = 27

4 2 = 16*2=32

5 1 = 25*1=25

6 0 = 0 in formula

32 is the maximum value of the formula when u+v=6

y = 4/x so f = x² + 64/x³

df/dx = 2x -192/x^4

For stationary points x^5 = 96 so x= 2{(3)^(1/5)}

d²f/dx² = 2 + 768/x^5 which is positive at above point so point is a minimum.

Min f = x²(1+64/x^5) = x²(1+64/96) = (20/3){(3)^(2/5)}

Plot objective on WolframAlpha. This is global min in 1st quadrant.

Again put v=6-u so f = 6u²-u³

For stationary points df/du = 12u-3u²=0 so u=0, u=4

d²f/du² = 12-6u

So u=0, v=6, f=0 is a min : u=4, v=2, f=32 is a max :

Local max/min, but that's all these methods expect to find.