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precalculus project

alex zamora rafael gutierrez angel gonzalez martin perez

Carlos Gutierrez

on 19 April 2010

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Transcript of precalculus project

Math 5 final proyect Let's say a farmer tries to optimize the way he´s farm works To do so he relays on differential calculus In calculus; Optimization is defined as seeking for the maximum or minimum points using a scalar, real-valued objective function The farmer wishes to start by drawing parameter for every vegetable, A square for the carrots & a Circle for the onions. however, he only has enough money for 120 meters wire to use on both perimeters due to his capital of 12000$ and the cost of the fence being 100$/m We can use optimization to find the area of each crop, first we define the formulas.
The area of the square is given by: l^2 which is x^2
The area of the circle is given by: π*r^2
Therefore the ecuation of the total ares is given by: x^2+πr^2
The perimeter on the other hand is given by: 4x*2πr; which is the sume of the perimeter of the square 4x plus the perimeter of the circle 2πr. If we solve the perimeter ecuation for "r" we obtain a function if variable "x" tha we can substitute on the function of Area. 120= 4x + 2πr
r=(2(30x-x))/π Then we substitute "r" in the original funciton.
Then simplify and derivate:
A= x^2+(4(900-30x+x^2))/π
A= (x^2 (π+4)+3600+120x)/π
A= 1/π(x^2 (π+4)+3600-120x)
A'=1/π(2(x+4)x-120) Afterwards we have to equal the expression to zero(0) and solve for x.
1/π (2(π+4)x-120)=0
x=60/((π+4) )
Then we can substitute further by substituting "x" nad obtain our area.
((60/(π+4))^2 (π+4)+3600-120(60/(π+4)))/π=A
(60/(π+4))^2 (π+4)+3600-120(60/(π+4))=A
There now you substitute the value in the totl are function to obtain how much each figure will represent. The farmer also wants a model of open boxes with square base, that have a surfface area of 20cm2 . He can afford a couple of boxes with 80$ at 40$ each. The equation of Volume of the box is given by:
And the primary equation to optimize the size of the box is:
S= x^2+4xh=20 the sume of the area of the base plus the area of the sides.
You solve the equation for "h"and get a funtionc just like we did with "r" in the last problem.

After you obtain the "h" you need to substitute it in the function for volume V=x^2 h.
V= x^2 ((20-x^2)/4x)
V=((20-x^2 ) x^2)/4x

We must define the possible domain:
which sould be more than zero and less than the value we are looking for:
v ≥0 A=x^2
x ≥0 and x≤√20
Once that we define the critical numbers:
Now we can finish by substituting the value in the equation for volume:
V=12.9 – 4.29

And we're done. You obtained the maximum volume of the boxes.

Carlos Rafael Gutierrez Andrade
Alex Zamora
Angel Gonzalez
Martin Perez
Full transcript