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# Calculus - Optimizing Project

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## Ramez A.

on 6 March 2015

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#### Transcript of Calculus - Optimizing Project

Maximizing Can #1
"Vegetable Soup"
SO....What is Optimization?
~~> Optimization is the procedure to determine the
best possible outcome
of a given situation.

~~> For a function, it's the process of finding the critical values which are the zeros and max and min.

~~> In this case, we are finding the max points on a Radius-Volume graph.

~~> In other words, it is your
Critical Points.
Once Again, Surface Area...
Surface Area of a Cylinder can be found using this formula. This formula is derived by simply adding the surface area of 2 top and bottom circles and the sides.
What do we NEED?
What do we know? What don't we know?
According to this formula, we must get the
and the
height
of the can in order to get Surface Area.
: The most accurate method to attain the radius is using the
Circumference of the can which can be found by wrapping a piece of thread around the can.
Now we can rearrange for the Radius:
Breaking It Down!
Volume Optimized
CONCLUSION
How To Get The Most Of Your Canned Foods Using CALCULUS!
THANK YOU!
First we have to see how much material is used in total to make each canned item. This is called the
Surface Area.
Then we can use optimization to find our maximum dimensions!

HEIGHT:
The height can easily be found using a ruler and measuring the height of the can.
(1) Circumference measured around the
can & equals to 20.7 cm.

C = 2πr
20.7 =2πr
r = 20.7/2π

Finding Height:
Using a ruler, Height comes out to be 9.8 cm
Surface Area:
S.A = 2πr² + 2πrh

= 2π(20.7/2π)² + 2π(20.7/2π)(9.8)

= 271.0563016 cm²
Maximizing Can #2
"Coconut Milk"
(1) Measured around the
can & Circumference equals to
20.7 cm.

C = 2πr
23.7 =2πr
r = 23.7/2π

Finding Height:
Using a ruler, Height comes out to be 11 cm
Surface Area:
S.A = 2πr² + 2πrh

= 2π(23.7/2π)² + 2π(23.7/2π)(11)

= 350.09574 cm²
Maximizing Can #3
"Solid White Tuna"
(1) Measured around the
can & Circumference equals to
27.0 cm.

Finding Height:
Using a ruler, Height comes out to be 4.0 cm
Surface Area:

S.A = 2πr² + 2πrh

= 2π(27/2π)² + 2π(27/2π)(4)

= 224 cm²
Now that we know what the Surface Area is, we can start the procedure to maximize these cans. We do this by getting the derivative of the volume of these cylinder-shaped cans using...
PROBLEM!
...We have two unknown variables. What can we do? Isolate one in terms of the other. Great Thinking!
Since we know what the SA is, we can plug that back in, and find R (The Radius) in terms of H (The Height). Then we are able to get an equation for the volume.
Maximizing Can #1
"Vegetable Soup"
Solving For H In Terms Of R
Since We Know What SA is:

S.A = 2πr² + 2πrh
(Plug in &
Rearrange)
271.0563016 - 2πr² = 2πrh
(Isolate)
H = ( 271.0563016 – 2πr²) / (2πr)
Using this new equation for H in terms of R,
we can now get an equation for the volume of this can
and derive it!

V(R) = πr²h
*Sub in H
= πr²[( 271.0563016 – 2πr²) / (2πr)]
*Simplify by crossing out. (One of the crucial tricky algebraic tricks .. saves alot of time)
= ( 271.0563016 r – 2πr³) / 2
Maximizing Can #2
"Coconut Milk"
Solving For H In Terms Of R
Since We Know What SA is:

S.A = 2πr² + 2πrh
(
Rearrange)
350.09574 - 2πr² = 2πrh
(
Isolate)
H = (350.09574 – 2πr^2) / (2πr)
Using this new equation for H in terms of R,
we can now get an equation for the volume of this can
and derive it!

V(R) = πr²h

= πr²[(350.09574 – 2πr²) / (2πr)]

= (350.09574r – 2πr³) / 2
Maximizing Can #3
"Solid White Tuna"
Solving For H In Terms Of R
Since We Know What SA is:

S.A = 2πr² + 2πrh

224 - 2πr² = 2πrh

H = (224 – 2πr²) / (2πr)
Using this new equation for H in terms of R,
we can now get an equation for the volume of this can
and derive it!

V(R) = πr²h

= πr²[(224 – 2πr²) / (2πr)]

= (224r – 2πr³) / 2
Now that we have a function expressing the Volume of each can, we can get the derivative of these functions, which is in other words the "
Slope
" of the function showing all x values which is your radius, with the corresponding y values which is your Volume! Then you can get the
Critical Points

by equating the function to zero and isolating for the Radius, which allows us to find our local max and min Radius points because it looks for wherever you have a horizontal tangent line. You automatically ignore the
negative value
since there is

no

Can #1 ~ Vegetable Soup!
V(R)
= ( 271.0563016 r – 2πr³) / 2
(Get Derivative Using Calculus Knowledge)
V'(R) = ( 271.0563016 – 6πr²) / 2
(Equal To Zero)
0 = 271.0563016 – 6πr²
(Isolate for R)
6πr²= 271.0563016
R² = 271.0563016 / 6π
(Square root; Ignore Negative!)
R= √( 271.0563016 / 6π)
R = 3.792094859 cm

Can #2 ~ Coconut Milk!
(Get Derivative Using Adv' Functions Knowledge)
V'(r) = (350.09574 – 6πr²) / 2
(Equate To Zero)
0 = 350.09574 – 6πr²
(Isolate for R))
6πr²= 350.09574

R² = 350.09574 / 6π
(Square Root; Ignore Negative)
R² = √(350.09574 / 6π)

Therefore your Max R = 4.309658439 cm

Can #3 ~ Solid White Tuna!
(Get Derivative Using Adv' Functions Knowledge)
V'(r) = (224 – 6πr²) / 2
(Equate to Zero)
0 = 224 – 6πr²
(Isolate for R)
6πr²= 224

R²= 224 / 6π
(Square Root' Ignore Negative)
R = √(224 / 6π)

Therefore your Max R = 3.45 cm
We now know our Maximum Radius, but can we figure out Maximum Height? ...yes! Using the Surface Area equation in which we isolated for the Height ...
S.A = 2πr² + 2πrh
H = (S.A - 2πr²) / (2πr)
Can #1 ~ Vegetable Soup
(Plug in Max R value)
Max Height = [ 271.0563016 – 2π(3.792094859)²] / [2π(3.792094859)]
= 7.582902431 cm

Can #2 ~ Coconut Milk
Max Height = [350.09574 – 2π(4.309658439)²] / [2π(4.309658439)]
= 8.619316877 cm

Can #3 ~ Solid White Tuna
Max Height = [224 – 2π(3.45)²] / [2π(3.45)]
= 6.87 cm

What's the next step?...
Max Volume!
Since we have our Maximum Radius and Height, we can now able to find our new-and-improved optimized volume using; & compare it to our initial volume using our original height and radius!
New&Improved Optimized Volume VS Original Volume!
For Vegetable Soup
Original V = π(9.8)(20.7/2π)²
= 334.1618777 cm³

Max V =π(3.792094859)²(7.582902431)
= 342.5655814 cm³

For Coconut Milk
Original V = π(11)(23.7/2π)²
= 491.6765699 cm³

Max V = (4.309658439)²(8.619316877)
= 502. 9310201 cm³
For Solid White Tuna

Original V = π(4) (27/2π)²
= 232.4 cm³

Max V =(3.45)²(6.87)
= 257.5 cm³
How Close Is The Actual Can To Being The "
Highest Volume
" It Could Be?
In other words, does that metal really
enclose the most volume it could? Using this formula, we can know exactly how close was the can's volume to being maximized.

Volume Of Can
334.1618777 491.6765699 224
------------------------
------------------- -------------------- -------
Optimized Volume
342.5655817 502.9310201 257.5

=97.55% =97.76% =86.99%
"CAN You Fit More?"
Optimized
Actual!
Full transcript