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Transcript of precalculus project
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 (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.
Then we can substitute further by substituting "x" nad obtain our area.
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