"The quantity, q, of a product manufactured depends on the number of workers, W, and the amount of capital invested, K, and is given by

q = 6 * W^(3/4) * K^(1/4)

Labor costs are $10 per worker and capital costs are $20 per unit, and the budget is $2000.

What are the optimum number of workers and the optimum number of units of capital?"

Thank you in advance for any help!

2000=(q)(10+20)

q=2000/30=667

dq=6*3/4*w^-.25 K^.25 + 6w^.75 * .25K^-.75 =0

18/4*K=6/4 w
3K=w

but from the original equation
667=6 w^.75 * (w/3)^.25
667=6 w (1/3)^.25

and you have w, then go back and k is 3 times that.

check my work, it is easy to err when typing these

12

To determine the optimum number of workers and units of capital, we need to maximize the quantity, q, while staying within the given budget.

Let's start by multiplying the labor cost by the number of workers to get the total labor cost:
labor cost = $10 * W

Next, multiply the capital cost by the units of capital to get the total capital cost:
capital cost = $20 * K

Since the budget is $2000, we can write the budget constraint as:
total cost = labor cost + capital cost ≤ $2000

Now, substitute the expressions for labor cost and capital cost into the budget constraint equation:
$10 * W + $20 * K ≤ $2000

To find the optimum number of workers and units of capital, we need to maximize the quantity q. However, we can simplify the optimization problem by defining a new variable, C, as the total cost:

C = $10 * W + $20 * K

Now, the problem can be stated as:
Maximize q = 6 * W^(3/4) * K^(1/4)
subject to the constraint C = $10 * W + $20 * K ≤ $2000

To solve this optimization problem, we can use the Lagrange multipliers method. The Lagrangian function is given by:

L = q - λ(C - $2000)

Differentiate the Lagrangian function with respect to W, K, and λ, and set the derivatives equal to zero to find the critical points.

Differentiating L with respect to W:
∂L/∂W = (3/2) * (6^(1/4)) * (K^(1/4)) * (W^(-1/4)) - λ * 10 = 0

Differentiating L with respect to K:
∂L/∂K = (1/4) * (6^(3/4)) * (W^(3/4)) * (K^(-3/4)) - λ * 20 = 0

Taking the derivative of C with respect to λ:
∂C/∂λ = C - $2000 = 0

Simplifying the equations and solving them simultaneously will give us the values of W, K, and λ. However, this requires more calculation, so let's solve it numerically.

Using software like Excel, MATLAB, or an online optimization tool, you can setup the objective function q and the constraint C and solve for the optimal values of W and K that satisfy the budget constraint.

The optimal values of W and K will give you the optimum number of workers and the optimum number of units of capital, respectively, to maximize the quantity q within the given budget.

To determine the optimum number of workers and the optimum number of units of capital, we need to maximize the quantity of the product subject to the budget constraint.

Let's start by introducing the budget constraint. The total cost of labor and capital cannot exceed the budget ($2000). We can express this as:

10W + 20K ≤ 2000 ---- (Equation 1)

Next, we need to find the optimum values of W and K that will maximize the quantity of the product.

To find the optimum values, we can use the method of partial derivatives.

First, calculate the partial derivative of q with respect to W, holding K constant:

∂q/∂W = (3/4) * 6 * W^(-1/4) * K^(1/4) ---- (Equation 2)

Next, calculate the partial derivative of q with respect to K, holding W constant:

∂q/∂K = (1/4) * 6 * W^(3/4) * K^(-3/4) ---- (Equation 3)

To find the optimum values of W and K, we need to find the values that make both partial derivatives equal to zero.

Setting Equation 2 equal to zero:

(3/4) * 6 * W^(-1/4) * K^(1/4) = 0

Simplifying, we have:

W^(-1/4) * K^(1/4) = 0

Since W and K must be positive, W^(-1/4) cannot be zero. Therefore, K must be zero for Equation 2 to hold true. However, having K=0 would contradict the budget constraint in Equation 1, so K cannot be zero.

Therefore, we can conclude that there is no critical point for W using Equation 2.

Setting Equation 3 equal to zero:

(1/4) * 6 * W^(3/4) * K^(-3/4) = 0

Simplifying, we have:

W^(3/4) * K^(-3/4) = 0

Again, since W and K must be positive, K^(-3/4) cannot be zero. Therefore, W must be zero for Equation 3 to hold true. However, having W=0 would contradict the budget constraint in Equation 1, so W cannot be zero.

Therefore, we can conclude that there is no critical point for K using Equation 3.

Since we have no critical points, we cannot find the optimum values for W and K using the method of partial derivatives. This implies that the given production function and budget constraint do not allow for a unique maximum quantity of the product.

In this case, it is likely that the maximum quantity of the product occurs at an endpoint of the feasible region defined by the budget constraint. To find this, we need to consider the boundaries of the feasible set defined by Equation 1.

Let's solve Equation 1 for K:

20K ≤ 2000 - 10W

K ≤ (2000 - 10W)/20

Since K must be a non-negative integer, we can consider values for K that satisfy this inequality within the range of the budget constraint when W varies.

For example, when W = 0, K can take any non-negative integer value. When W = 1, K ≤ (2000 - 10)/20 = 1900/20 = 95. And so on.

To determine the optimum number of workers and the optimum number of units of capital, we need more information about the specific values of W and K within the feasible region.