a builder plans to construct a three-story building with a rectangular floor pla

Question

a builder plans to construct a three-story building with a rectangular floor plan. the cost of the building, which depends on the dimensions of the floor plan, is given by xy+30x+20y+474,000 where x and y are, respectively, the length and width of the rectangular floors. What length and width should be used if the building has a budget of 500,000 and should have maximum area on each floor.

I am looking for guidence not answer. I want to know a stratagy to solve these type of problems and how you would start this one.

Explanation / Answer

There are two basic methods. One is to solve for x in terms of y using the constraint and substitute this into the equation you are maximizing (referred to as the objective function), then maximize over y. The other way is to use Lagrange multiplier theory, which involves dealing with the constraints and objective function at the same time.

In this problem, we are maximizing area. As the building is rectangular, this is simply xy.

As the budget is 500000, and the cost is xy + 30x + 20y + 474000,

Our constraint is xy + 30x + 20y + 474000 = 500000, or

xy + 30x + 20y = 26000

Then, x(30+y) = 26000 - 20 y
x = (26000 - 20 y)/(30+y)
Thus, if we wanted, we could substitute this into xy and maximize over y
xy = (26000 - 20 y)/(30+y) y =
(26000 y - 20 y2)/(30+y)

Taking the derivative, we get -((26000 - 40 y)(30 +y) - (26000 y - 20 y2))/(30+y)2

We set this equal to 0 by setting the numerator equal to 0

((26000 - 40 y)(30 +y) - (26000 y - 20 y2)) =

780000 + 24800 y - 40 y 2 - 26000 y + 20 y 2 =

780000 - 1200 y - 20 y 2 = 0

Dividing by -20

y2 + 60 y - 39000 = 0

y = -30 + 10399

x = (26000 - 20 y)/(30+y) =

(26000 - 20 (-30+10399))/(30+(-30+10399) =

26600 -200399/10399 = (as 26600 = 133 * 2 * 100 = 399 * 100 * 2/3)

-20 + 20/3399

The other way to solve this problem is to use LaGrange Multipliers.

Then, l(x,y) = xy + (xy + 30x + 20y - 26000) (note the maximization function and constraint function)

and you calculate the gradient -

(y + y + 30, x + x + 20) and set equal to 0.

Solving for , we get y = -y - 30, or = -1 -30/y

Also, x = -x - 20, or = -1 - 20/x

Setting the two equations equal, we get -1 - 30/y = -1 - 20/x, or 30/y = 20/x, or x = 2/3 y

Substituting back into the constraint equation xy + 30x + 20y - 26000 = 0, we get

2/3yy + 30(2/3y) + 20y - 26000 = 0, or

2/3y 2 + 40y - 26000 = 0

Multiply by 3/2, and we get

y2 + 60 y - 39000 = 0

This matches the equation above, so we get the same solution

x = 2/3y = -20 + 20/3 399

We get the same answer both ways.

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