I don't know to do this 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?

2 answers

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
(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 ...
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