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# precalculus project

alex zamora rafael gutierrez angel gonzalez martin perez
by

## 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
(120-4x)/2π=r
r=(2(30x-x))/π Then we substitute "r" in the original funciton.
A=x^2+(2(30-x)/π)
Then simplify and derivate:
A=x^2+(4(30-x)^2)/π
A= x^2+(4(900-30x+x^2))/π
A=x^2+(3600+120x+x^2)/π
A=(πx^2+3600+120x+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
2(π+4)x=120
(π+4)x=60
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
3600+3600-120(60/(π+4))=A
7120-120(60/(π+4))=A
7120-1008.17=A
A=6111.83m^2
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:
V=x^2h
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.
x^2+4xh=20
4xh=20-x^2
h=(20-x^2)/4x

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
V=(20x^2-x^4)/x
V=5x-x^3/4

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:
F’(x)=5-(3x^2)/4
5-(3x^2)/4=0
-(3x^2)/4=-5
3x2=-20
x2=6.67
x=√6.67
x=±2.58
x=2.58cm
Now we can finish by substituting the value in the equation for volume:
5(2.58)-((2.58))/4
V=12.9 – 4.29
V=8.61cm3

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