3 . The production of housing has the following form:

H = number of housing units = 4(L + 10)1/2 + 80 (M + 3)3/4, where L is land and M is
materials. One unit of L costs 100 and 1 unit of M costs 12.

Find the cheapest ways to produce 1000 housing units, 2000 housing units, and 5000
housing units.

To find the cheapest ways to produce 1000, 2000, and 5000 housing units, we need to calculate the costs for different combinations of land and materials.

Given:
- L (land) costs 100 per unit.
- M (materials) cost 12 per unit.
- The production function for housing units is given by H = 4(L+10)^(1/2) + 80(M+3)^(3/4).

Let's calculate the costs for each scenario:

1. 1000 housing units:
We need to find the combination of L and M that minimizes the cost for producing 1000 housing units.

Let's assume x units of L and y units of M. To produce 1000 housing units, we have the following equation:

1000 = 4(x + 10)^(1/2) + 80(y + 3)^(3/4) (Equation 1)

The cost (C) can be calculated as follows:
C = 100x (for land cost) + 12y (for material cost)

To find the cheapest way, we need to minimize the cost (C) by solving Equation 1.

2. 2000 housing units:
Similarly, for 2000 housing units, we have:

2000 = 4(x + 10)^(1/2) + 80(y + 3)^(3/4) (Equation 2)

3. 5000 housing units:
And for 5000 housing units:

5000 = 4(x + 10)^(1/2) + 80(y + 3)^(3/4) (Equation 3)

To find the cheapest combinations, we will solve these equations one by one.

Step 1: Solve Equation 1 for 1000 housing units.

Let's rearrange Equation 1:

4(x + 10)^(1/2) = 1000 - 80(y + 3)^(3/4)

Square both sides of the equation:

16(x + 10) = (1000 - 80(y + 3)^(3/4))^2

Simplify:

16x + 160 = (1000 - 80(y + 3)^(3/4))^2 (Equation 4)

Now, we need to minimize the cost (C) by finding the combination of x and y that satisfies Equation 4.

Step 2: Solve Equation 2 for 2000 housing units.

Proceeding similarly, for 2000 housing units, we have:

16x + 160 = (2000 - 80(y + 3)^(3/4))^2 (Equation 5)

Step 3: Solve Equation 3 for 5000 housing units.

For 5000 housing units:

16x + 160 = (5000 - 80(y + 3)^(3/4))^2 (Equation 6)

Now, we need to solve Equations 4, 5, and 6 to find the values of x and y that satisfy each equation.

To find the cheapest ways to produce a certain number of housing units, we need to consider the costs of land (L) and materials (M).

Given:

Cost of 1 unit of L = $100
Cost of 1 unit of M = $12

Production function:

H = 4(L + 10)^(1/2) + 80(M + 3)^(3/4)

Now, let's calculate the costs for producing different quantities of housing units.

1. For 1000 housing units:
We want to minimize the cost C = (number of L units) * (cost of 1 L unit) + (number of M units) * (cost of 1 M unit), subject to the constraint H = 1000.

Substitute H = 1000 into the production function:

1000 = 4(L + 10)^(1/2) + 80(M + 3)^(3/4)

Now, we have an equation with two variables (L and M). To find the values of L and M that minimize the cost C, we need to use an optimization techniques such as calculus or a numerical method. For simplicity, let's use calculus:

Differentiate both sides of the equation with respect to L:

d/dL(1000) = d/dL(4(L + 10)^(1/2) + 80(M + 3)^(3/4))

0 = 2(L + 10)^(-1/2) * 4 + 0
0 = 8(L + 10)^(-1/2)

Solve for L:

(L + 10)^(-1/2) = 0
L + 10 = ∞
L = ∞ - 10
L = ∞

Since we can't obtain a finite value for L, it means that we have no constraint on L and can set it to any value. Let's set L = 0.

Now, substitute L = 0 into the production function:

1000 = 4(0 + 10)^(1/2) + 80(M + 3)^(3/4)
1000 = 4(10)^(1/2) + 80(M + 3)^(3/4)

Let's solve for M:

996 = 80(M + 3)^(3/4)
(M + 3)^(3/4) = 996/80
(M + 3)^(3/4) = 12.45

Raise both sides to the power of 4/3:

M + 3 = (12.45)^(4/3)
M + 3 = 24.14
M = 24.14 - 3
M = 21.14

Therefore, the cheapest way to produce 1000 housing units is by using 0 units of land (L) and 21.14 units of materials (M). The total cost will be:

Cost = (number of L units) * (cost of 1 L unit) + (number of M units) * (cost of 1 M unit)
Cost = 0 * 100 + 21.14 * 12
Cost ≈ $254.88

2. For 2000 housing units:
We can follow the same steps as above, only substituting H = 2000 into the production function and solving for L and M.

3. For 5000 housing units:
Again, follow the same steps as above, but substitute H = 5000 into the production function and solve for L and M.