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?
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 ...
a) To find the numbers that maximize the product of the first number and the square of the second number, we can use the method of optimization.
Let's assume the first number is x and the second number is y.
The problem states that the two numbers add up to 6, so we have the equation:
x + y = 6
We want to find the maximum value of the product, which is given by:
P = xy^2
To solve this problem, we can use the substitution method.
From the equation x + y = 6, we can solve for x in terms of y:
x = 6 - y
Substituting this expression for x in the equation for P, we get:
P = (6 - y)y^2
Expanding and simplifying, we have:
P = 6y^2 - y^3
To find the maximum value of P, we need to find the critical points, where the derivative of P with respect to y is zero.
Differentiating P with respect to y, we get:
dP/dy = 12y - 3y^2
Setting this derivative equal to zero, we have:
12y - 3y^2 = 0
Factorizing this equation, we get:
3y(4 - y) = 0
Setting each factor equal to zero, we have:
3y = 0 or 4 - y = 0
From the first equation, we get y = 0.
From the second equation, we get y = 4.
Now, we need to test these critical points to see if they correspond to a maximum value.
For y = 0, we have x + y = 6, which implies x = 6. Therefore, one of the numbers is 6 and the other is 0.
For y = 4, we have x + y = 6, which implies x = 2. Therefore, one of the numbers is 2 and the other is 4.
To determine which of these values gives the maximum product, we can compare the values of P.
For y = 0, P = (6 - 0) * 0^2 = 0
For y = 4, P = (6 - 4) * 4^2 = 32
Therefore, the maximum product is obtained when one number is 2 and the other number is 4.
b) To determine the optimum dimensions of the corral that minimize the amount of fencing required, we can use the method of optimization.
Let's assume the length of the corral is x metres and the width is y metres.
The problem states that the internal fence splits the corral into two equal areas, so the total area of the corral can be expressed as:
A = xy
The total perimeter of the corral is given by:
P = 2x + 2y
We want to minimize the amount of fencing required, which is given by the perimeter. Therefore, we need to minimize the function P.
Since we have two variables (x and y), we need to express one variable in terms of the other using a constraint.
From the given information, we know that the corral's total area is 9600 square metres:
xy = 9600
We can solve this equation for x in terms of y:
x = 9600/y
Substituting this expression for x in the equation for P, we get:
P = 2(9600/y) + 2y
Simplifying, we have:
P = 19200/y + 2y
To find the minimum value of P, we need to find the critical points, where the derivative of P with respect to y is zero.
Differentiating P with respect to y, we get:
dP/dy = -19200/y^2 + 2
Setting this derivative equal to zero, we have:
-19200/y^2 + 2 = 0
Simplifying, we have:
-19200 + 2y^2 = 0
Rearranging, we have:
2y^2 = 19200
y^2 = 9600
Taking the square root of both sides, we have:
y = ± 96
Since y represents the width of the corral, a negative value does not make sense, so we have:
y = 96
Substituting this value for y into the equation for x, we have:
x = 9600/96 = 100
Therefore, the optimum dimensions of the corral that minimize the amount of fencing required are a length of 100 metres and a width of 96 metres.
To solve optimization problems, we need to set up an objective function and constraints, and then find the maximum or minimum value of the objective function within those constraints.
a) To find the two numbers that maximize the product of the first number and the square of the second number:
Let's call the first number x and the second number y.
We know that the two numbers add up to 6, so we have the constraint: x + y = 6.
Now, we need to define the objective function that we want to maximize. In this case, it's the product of x and y squared, which can be written as f(x, y) = xy^2.
To solve this optimization problem, we need to use the method of Lagrange multipliers:
1. Set up the Lagrangian function L(x, y, λ) = xy^2 + λ(x + y - 6).
2. Take partial derivatives of L with respect to x, y, and λ, and set them equal to zero.
∂L/∂x = y^2 + λ = 0
∂L/∂y = 2xy + λ = 0
∂L/∂λ = x + y - 6 = 0
3. Solve this system of equations to find the values of x, y, and λ.
From the first equation, we get y^2 = -λ. Since y is greater than zero, λ must be negative.
From the second equation, we get x = -λ/2y.
Plugging these into the third equation, we get -λ/2y + y - 6 = 0.
Simplifying this equation, we have -λ + 2y^2 - 12y = 0.
Since λ is negative, we can multiply the equation by -1 to eliminate the negative sign: λ - 2y^2 + 12y = 0.
We can solve this quadratic equation for y, and then find x using x = -λ/2y.
4. Once we have the values of x and y, we can calculate the value of the objective function f(x, y) = xy^2, which will give us the maximum product.
b) To find the optimum dimensions of the horse corral that minimize the amount of fencing required:
Let's call the length of the corral x and the width of the corral y.
We know that there is a straight internal fence that splits the corral into 2 equal areas, so the length of each half of the corral is x/2.
The total area of the corral is given as 9600 square meters, so we have the constraint: x * y = 9600.
We want to minimize the amount of fencing required, which is the perimeter of the corral. The perimeter is given by P = x + 2y + x/2.
To solve this optimization problem, we can express the perimeter in terms of a single variable and then minimize that expression:
1. Rewrite the perimeter equation as a function of a single variable:
P = x + 2y + x/2
= 2y + x(1 + 1/2)
= 2y + x(3/2)
= 2y + (3/2)x.
2. Replace one variable in terms of the other using the area constraint:
x * y = 9600
y = 9600 / x
3. Substitute the expression for y back into the perimeter equation, so we have the perimeter only in terms of x:
P = 2(9600 / x) + (3/2)x.
4. We now have a single-variable function for the perimeter, P(x). Differentiate P with respect to x, and set the derivative equal to zero to find the critical point(s).
5. Once we have the critical point(s), we can evaluate the perimeter at those points and find the one that gives the smallest value. This value corresponds to the optimum dimensions of the corral.