"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

radius

and the

height

of the can in order to get Surface Area.

RADIUS

: 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.

Finding Radius:

(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"

Finding Radius:

(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"

Finding Radius:

(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

negative Radius.

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)

Therefore your Max

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!