I don't know to do this question.

Question

5. Solve the following optimization problems:

a) Two numbers greater than zero add up to 6. Find the numbers so that the product of the first number and the
square of the second number is maximum.

b) A horse corral is rectangular, with fencing around the perimeter. Also, there is a straight internal fence, parallel to
the sides. The internal fence splits the corral into 2 equal areas. The total area of the corral is 9600 square metres.
The owner wishes to minimize the amount of fencing required. What are the optimum dimensions of the corral?

Answer 1

Let the two numbers by x and y
x+y = 6 ---> y = 6-x

product = yx^2
= x^2(6-x) =6x^2 - x^3
d(product)/dx = 12x - 3x^2
for a max/min, 12x - 3x^2 = 0
3x(4 - x^2) = 0
x = 0 , or x = ± 2
but the numbers are to be positive, so x = 2, and y = 4
the two numbers are 2 and 4, with 4 as the number that would be squared

2nd question:
let the shorter side be x and the longer side be y
Assume that the internal side is parallel to the shorter side,
so you have xy = 9600 or y = 9600/x

minimum cost ---> minimum perimeter
= 3x + 2y
= 3x + 19200/x
d(perimeter)/dx = 3 - 19200/x^2 = 0 for a min
3x^2 = 19200
x^2 = 6400
x = ± 80

so the short side is 80 and the longer side is 9600/80 = 120 metres

Answer 2

(a)
so, we have z = xy^2 = (6-y)y^2 = 6y^2-y^3
dz/dy = 12y-3y^2 = 3y(4-y)
dz/dy=0 when y=0 or 4
But we know that y>0, so we must have (x,y) = (2,4) and max z is thus 32

(b)
Let the corral have width=2x and length=y
Let the internal fence be parallel to the length.
So, we have 2xy=9600, so xy=4800
we want to minimize p=2(2x+y) = 2(2x+4800/x) = 4(x + 2400/x)
dp/dx = 4(1-2400/x^2)
so, dp/dx=0 when x^2=2400
x = 20√6
y = 4800/x = 240/√6 = 40√6
So the perimeter p(20√6,40√6) = 80√6

As with all of these problems, maximum area (or minimum perimeter) is achieved when the available fencing is divided equally between lengths and widths.

I am assuming that this homework assignment was not intended to be solved using Lagrange multipliers ...