Very Basic Maths Assignment (sorry i'm a newbie) based on profit and graphs

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The discussion revolves around a linear programming problem where a builder must decide how many luxury and standard houses to construct on a 6000m2 plot while adhering to specific constraints. The builder is required to build at least 5 luxury and 10 standard houses, with a maximum of 30 houses total. The area required for each type of house is 300m2 for luxury and 150m2 for standard, and the profit per house is $12,000 for luxury and $8,000 for standard. The optimal solution involves formulating the problem using inequalities and can be solved graphically or with spreadsheet tools like Excel Solver. Ultimately, the maximum profit achievable is $280,000 by building 10 luxury houses and 20 standard houses.
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Homework Statement



Firstly I'm sorry to bother all of you that will find this question really basic but I haven't studied Maths since GCSE 7 years ago and I am now doing a degree in Engineering without A Level Maths so any help would be great...I need to work out the algebra and the main issue is how to plot it on a graph!? I'm guessing I need to use y=mx + c but I'm not sure how to do that on excel.

A builder has a plot of land available on which he can build either luxury of standard houses.
He decides to build at least 5 luxury and 10 standard houses. Planning restrictions limit him
to no more than 30 houses altogether. A luxury house requires 300m2
of land and a standard
house 150m2
. The plot is 6000m2
. Profit is $12; 000 per luxury house and $8000 per standard
house. How many of each type should he build to maximise his profit.


Thank you anyone that can help!

Adam
 
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ok so here is what I have worked out so far...

Luxury Houses = x
Basic Houses = y

(300x + 5y) + (150y + 10x) = 6000m2 i'm not too sure if this is correct

I also came to the conclusion for graph purposes the answer might be y= -0.5x + 10 ?

I also know the most profit is £280,000 with 10 luxury houses and 20 basic.
 
You need a change of variables. you know you need a certain number of lux and std minimum and you know the maximum houses so subtract the mins from the max to obtain something along the lines of:

30-10 -5 = x+y => y = a function of x

you can then replace all y with the function of x and can solve this 2 variable question as a single variable. Remember to subtract the area of the min requirements from your total area btw.
 
If the builder is going to build at least 5 luxury and 10 standard homes, then calculate the amount of profit obtained from their construction. From the total amount of land, subtract the area required by these 15 homes. For the remainder of the land area, find out the mix of luxury/standard homes which will fit that gives the most additional profit.
 
adste89 said:

Homework Statement



Firstly I'm sorry to bother all of you that will find this question really basic but I haven't studied Maths since GCSE 7 years ago and I am now doing a degree in Engineering without A Level Maths so any help would be great...I need to work out the algebra and the main issue is how to plot it on a graph!? I'm guessing I need to use y=mx + c but I'm not sure how to do that on excel.

A builder has a plot of land available on which he can build either luxury of standard houses.
He decides to build at least 5 luxury and 10 standard houses. Planning restrictions limit him
to no more than 30 houses altogether. A luxury house requires 300m2
of land and a standard
house 150m2
. The plot is 6000m2
. Profit is $12; 000 per luxury house and $8000 per standard
house. How many of each type should he build to maximise his profit.


Thank you anyone that can help!

Adam

This is a standard "linear programming" problem, and involves *inequalities*, not equations. Here is one formulation. Let L = number of luxury houses to build and S = number of standard houses to build. (No reason to use symbols x and y, when symbols L and S are so much more tied to their definitions.) Land used = 300L + 150S, and this must be <= 6000. Total number of houses = L + S, and this must be <= 30. Also: L >= 5 and S >= 10. Finally: profit = 12L + 8S (in $000,s). The final linear programming (LP) formulation is:
max 12L + 8S,
subject to
300L + 150S <= 6000
L + S <= 30
L >= 5, S >= 10

There are standard algorithms available for solving such problems manually, or you can submit it to a Solver tool in a spreadsheet (eg., the EXCEL Solver can deal with such problems having up to about 200-300 decision variables and about 200 inequalities and equations, together with simple bounds that are not part of the inequality count). However, simple 2-variable problems like this one can be tackled using a _graphical method_ because you have only two variables (L and S), so can plot the constraint region in a 2-dimensional plot and look where in the constraint region the 'objective function' 12L + 8 S is maximized. See, eg., http://math.tutorvista.com/algebra/graphical-method.html or any introductory Operations Research textbook.

RGV
 

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