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.

Carlos Rafael Gutierrez Andrade

Alex Zamora

Angel Gonzalez

Martin Perez

### Present Remotely

Send the link below via email or IM

CopyPresent to your audience

Start remote presentation- Invited audience members
**will follow you**as you navigate and present - People invited to a presentation
**do not need a Prezi account** - This link expires
**10 minutes**after you close the presentation - A maximum of
**30 users**can follow your presentation - Learn more about this feature in our knowledge base article

# precalculus project

alex zamora
rafael gutierrez
angel gonzalez
martin perez

by

Tweet