Find the minimum cost of producing 20,000 units of a product, where x is the number of units of labor (at$48 per unit) and y is the number of units of capital (at $36 per unit).
P(x,y)=100x^.6y^.4
To find the minimum cost of producing 20,000 units of a product, we need to minimize the cost function P(x, y) = 100x^0.6y^0.4, where x represents the units of labor and y represents the units of capital.
To minimize this cost function, we can use calculus. We need to find the partial derivatives of P(x, y) with respect to both x and y, and set them equal to zero:
∂P/∂x = 60x^(-0.4)y^(0.4) = 0
∂P/∂y = 40x^0.6y^(-0.6) = 0
Solving these equations will give us the critical points where the minimum cost occurs.
For the first equation, 60x^(-0.4)y^(0.4) = 0, either x=0 or y=0. However, since we are looking to produce 20,000 units, it's not realistic for either x or y to be zero. Therefore, we can rule out these solutions.
For the second equation, 40x^0.6y^(-0.6) = 0, we can rewrite it as y^(-0.6) = 0. We cannot solve for y since it would lead to a division by zero, so we need to consider points where y is very close to zero.
Now we need to check the endpoints of the feasible region. Since we want to produce 20,000 units, there is a constraint on the total output:
x * y = 20,000
We can use this constraint to solve for x in terms of y:
x = 20,000 / y
Substituting this into the cost function:
P(x, y) = 100 * (20,000 / y)^0.6 * y^0.4
Simplifying further:
P(y) = 100 * (20,000)^0.6 * y^(-0.24) * y^0.4
P(y) = 100 * (20,000)^0.6 * y^(0.4 - 0.24)
P(y) = C * y^0.16, where C is a constant
Since y cannot be zero, we can see that the minimum cost will occur when y is as close to zero as possible. Intuitively, this means we should focus more on labor (x) than capital (y) to minimize the cost.
In summary, to find the minimum cost of producing 20,000 units:
1. Solve the first equation, 60x^(-0.4)y^(0.4) = 0, to consider the critical points.
2. Consider the constraint x * y = 20,000 and solve for x in terms of y.
3. Plug the value of x into the cost function and simplify to obtain P(y).
4. Determine the value of y that minimizes P(y) by considering that y is as close to zero as possible.
5. Calculate the minimum cost using the corresponding value of y and the cost function.
I do not know if you have done Lagrange multipliers. Here's a non-linear application.
The basic ideas can be found in your textbook, or you can refer to a previous post:
http://www.jiskha.com/display.cgi?id=1291776389
We will apply the method here.
The production function, P(x,y) represents the number of units produced when the quantities of labour (x) and capital (y) are available, where
P(x,y)=100x0.6y0.4
This is our constraint equation, since the number of units produced must equal 20000, so
100x0.6y0.4-20000=0 ... (1)
For the resources, we would like to minimize the total cost, which can be represented by the cost function,
C(x,y)=48x+36y ....(2)
For C(x,y) to be a minimum, subject to the constraint equation (1), we form a new function,
Λ(x,y)=C(x,y)+λ(P(x,y)-20000) ....(3)
By partial differentiation of (3) with respect to x and y, we obtain equations
∂Λ/∂x
=48 - 60λx^(-0.4)y^(0.4) = 0 ...(4)
∂Λ/∂y
=48 - 40λx^(0.6)y^(-0.6) = 0 ... (5)
and finally the constraint equation (1)
100x0.6y0.4-20000=0 ... (6)
Solve algebraic equations (4), (5) and (6) for x,y and λ using trial and error, or other iterative methods, we obtain (this is the most tedious part of the problem):
x=209.648
y=186.354
λ does not enter into our calculations, but it equals 0.8385925
Total cost
=C(209.648,186.354)
=$16771.85
Check:
No. of units produced
=P(209.648,186.354)
=19999.99
Note: I did not make integral numbers of the resources, as it was not indicated in the question that these quantities are integers. In any case, the required quantity could be a multiple of 20000 in which case the fractional values could apply.