Question

a building contractor has 60 plots in which he can build a house per plot. He wishes to build houses of two types A and B. He wishes that houses of A type should be at least three times than those of B type. If he desires the profit of rs 65000 per house of type A and rs 80000 per house of type B to get maximum profit. How many houses of each type should he built.
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Answer 1

Answer:

Because of the unique nature of constructed facilities, it is almost imperative to have a separate price for each facility. The construction contract price includes the direct project cost including field supervision expenses plus the markup imposed by contractors for general overhead expenses and profit. The factors influencing a facility price will vary by type of facility and location as well. Within each of the major categories of construction such as residential housing, commercial buildings, industrial complexes and infrastructure, there are smaller segments which have very different environments with regard to price setting. However, all pricing arrangements have some common features in the form of the legal documents binding the owner and the supplier(s) of the facility. Without addressing special issues in various industry segments, the most common types of pricing arrangements can be described broadly to illustrate the basic principles.

Explanation:

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Answer 2

Answer:

Concept:

Profit and Loss

Explanation:

The decision variables are the number of house A and number of house B built to get maximum profit:

Let the number of house A built is $x_{1} ,$

Let the number of house B built is  $x_{2}$

STEP 2: Writing Objective function:

To maximize the profit:

$Z=65000 x_{1}+80000 x_{2}$

STEP 3: Writing Constraints:

$\begin{aligned}&x_{1}+x_{2} \leq 60 \\&x_{1} \geq 3 x_{2} \Rightarrow x_{1}-3 x_{2} \geq 0\end{aligned}$

and,

$x_{1} \geq 0, x_{2} \geq 0$

Result:

We have LP problem for the given situation is:

$\Max} Z=65000 x_{1}+80000 x_{2}$

Subject to,

$\begin{aligned}&x_{1}+x_{2} \leq 60 \\&x_{1}-3 x_{2} \geq 0\end{aligned}$

and,

$x_{1} \geq 0, x_{2} \geq 0$

We can solve this problem using the graphical method:

Hint to draw constraints:

1. To draw constraint $\times 1+\times 2 \leq 60 \rightarrow(1)$

Treat it as $\times 1+\times 2=60$

When $\times 1=0 then \times 2=?$

$\begin{aligned}&\Rightarrow(0)+\times 2=60 \\&\Rightarrow \times 2=60\end{aligned}$

When $\times 2=0 then \times 1=?$

$\begin{aligned}&\Rightarrow \times 1+(0)=60 \\&\Rightarrow \times 1=60 \\&\frac{\times 1060}{\times 2600}\end{aligned}$

2. To draw constraint$\times 1-3 \times 2 \geq 0 \rightarrow(2)$

Treat it as $\times 1-3 \times 2=0$

When $\times 1=0 then \times 2=?$

$\begin{aligned}&\Rightarrow(0)-3 \times 2=0 \\&\Rightarrow-3 \times 2=0\end{aligned}$

$\Rightarrow x 2=0-3=0 \\\\When \times 2=0 then \times 1=? \Rightarrow \times 1-3(0)=0 \Rightarrow \times 1=0 \frac{\times 100}{\times 200}$

The value of the objective function at each of these extreme points is as follows:

Extreme Point

The maximum value of the objective function Z=4125000 occurs at the extreme point (45,15).

Hence, the optimal solution to the given LP problem is : $x 1=45, \times 2=15 and \max Z=4125000 .$

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