The Last One

Over all the blog posts this will be the last one for the semester. So to wrap up all the blogs I’ve decided to end it all with something that threw me off on the last exam we took. It was a simple find the derivative problem but the appearance of it was daunting so I messed up. I’ll do a related example of the problem.

Find the derivative of

y = \frac {xex^x}{7x+29}

First we must remember the basic rules when it comes to dealing with derivatives of fractions.

f' = \frac {bt'-tb'}{b^2}

so we find b, b’, t, and t’ and plug them into the equation

b = 7x+29 and b'= 7

t = xe^x and t' = xe^x + e^x \cdot 1

Now to plug them into the equation…

f' =\frac {(7x + 29)(xe^x + e^x)-xe^x \cdot 7}{(7x + 29)^2}

Now we work it out…

= \frac {7x^2e^x+7xe^x+29xe^x+29e^x-7xe^x}{(7x+29)^2}

Now we simplify by removing parts that cancel each other out and we get our answer…

= \frac {(7x^2 + 29x +29)e^x}{(7x+29)^2}

I hope this will help people avoid any minor issues that derivatives may cause.

Boxes

It’s almost time for Christmas, you need to get the presents ready for wrapping. You have a box 20 in by 20 in that has a square cut  out of the corners.What is the maximum volume of the box for your Christmas present?

Blog box  

The equation for volume is V= l \cdot w \cdot h

The equation to find the maximized volume would be…

V = (20 - 2x) (20 - 2x) \cdot x

= (400 - 80x + 4x^2) \cdot x

= 400x - 80x^2 + 4x^3

Now we find the derivative…

V'= 400 - 160x + 12x^2

= 12x^2 - 160x + 400

Now we set it equal to zero…

12x^2 - 160x + 400= 0

We use the quadratic formula to help get the answer we need…

x= \frac {-b + or - \sqrt {b^2 - 4a \cdot c}}{2a}

x = \frac {-160 + or - \sqrt {-160^2 - 4(12)(400)}}{2(12)}

+ x = 10

- x = 3.33

Now we plug these two into the equation and see which one will work out…

V = (15 - 2(10))(15-2(10)) \cdot 10

= 250

V= (15- 2(3.33))(15-2(3.33)) \cdot 3.33

= 231.620

Since 250 is larger than 231.62o we find that the max volume of the box will be 250 in^3

 

 

Maximizing Pen Sizes

The Zoo has three new male Rhinoceri that don’t get along too well. In order to prevent violence, the zoo arranged three separate pen for each rhinoceros. Find the maximum area of the rhino pen with 8000 feet of fence.

rhino 1 for blog

 

We know that x \cdot y = 8000

So we use the equation P= 4x + 2y

P= 4x + 2 \frac {8000}{x}

Now we put it in a simpler mode to work with

=4x + 16000x^-1

Now for the derivative…

P'=4 - 16000x^-2

Now we solve for x…

4- \frac {16000}{x^2} =0

4x^2= 16000

x^2= 4000

x= \sqrt 4000

x= 63.246

Now we plug in x into the first equation

P= 4 \sqrt 4000 + 2 \frac {14000}{ \sqrt 4000}

So the total fence area is 2276.84 ft^2

 

 

 

 

 

More Optimization But With Multiple Columns

You will make a plastic rectangle that needs to be divided into 4 sections for specific paint colors. However, you know you need to make the area 30 square feet and you want to minimize the amount of plastic. how much plastic total would you need to create these columns of paint?

blog picture 1

 

Okay, so we know that the total area of the rectangle must be 30 ft^2

Our equation is P= x \cdot y= 30

So, P= 5x + 2y = 30

Knowing these two equations we know that P= \frac {30}{x}

So now lets set up the equation.

P= 5x + 2( \frac {30}{x} )

5x + 60x^-1

Now we find the derivative…

P'= 5-60x^-2

=5- \frac {60}{x^2}

Now we solve for x…

5x^2= 60

x^2=6

x= \sqrt{6}

x= 2.449

So now we plug in x into the equation P= 5x + 2 \frac {30}{x}

5 \sqrt{6} + 2 \frac {30}{ \sqrt {6}} = 134.722 ft^2

So the total area for the plastic is 134.722 ft^2

Practice With Simple Optimization

I have a few simple optimization problems for some good practice.

Find the maximum product of all pairs of non-negative numbers whose sum is 81.

We start by setting up the basic equation…

P= x(80-x)

Now we use algebra to solve for x

80x-x^2

Next we find the derivative…

P'= 80-2x

Now we set it to zero…

80-2x=0

80=2x

x=40

Now we insert 40 as x in the first equation…

P= 40(80-40)

P= 1600

Let’s try one more example.

Find the maximum product of all pairs of non-negative numbers whose sum is 42.

P= x(42-x)

Now to for the algebra…

42x-x^2

Now for the derivative…

P'= 42-2x

Now to solve for zero…

42-2x=o

42=2x

x=21

Now to plug it in…

P= 21(42-21)

P= 441

Cows Fear Water

Here’s a little word problem, a farmer wants to build a fence to keep his cattle all nice and organized. Since there is a river on his property and the cows are apparently hydrophobic he decided to go the less expensive route and build a fence with the river taking up one side. In order to keep the water-fearing cows well fed with grass, he has to have 220,000 square meters within the confines of the fence. What dimensions will minimize the amount of fence the farmer needs to not overpay for fencing?

Blog image 1

First off, you must recognize the constraints to this problem.

A = 220,000 m^2

x, y ≥ 0

x \cdot y = 220,000

Now we start with the equation F since we know that we have 2 x sides and one y for fencing.

F = 2x+y

Then we use what we know  from our constraints and determine that y= \frac {220,000}{x}

Next we plug that into our equation…

F= 2x + \frac {220,000}{x}

Now we change its form in preparation of finding the derivative…

F= 2x + 220,000x^{-1}

Now we find the derivative…

F'= 2 - 220,000x^{-2}

Then we change it back to its original form and set it equal to zero…

F'= 2- \frac {220,000}{x^2} = 0

For now we solve for x in the derivative.

2- \frac {220,000}{x^2}

2= \frac {220,000}{x^2}

2x^2 = 220,000

x^2 = 110,000

x= 331.662

Now we know that x = 331.662 m

To find y we divide 220,000 by x

y= \frac {220,ooo}{331.662}

y = 663.326

 

How to deal with Absolute Extrema

Okay, so I noticed that there was some issues understanding absolute extrema so I’m going to clear up any misconceptions that anyone may have.

We’ll start by looking at a graph of the function \frac {33x}{2x^2 + 7} + 3

This graph will help us determine the absolute extrema.

   With this given we know that \lim_{x \to \infty} f(x)= 3 

So with this information lets look for the highest and lowest points. However! It is vital to know what intervals that you are looking for the absolute extrema.

Graph for blog 2

For example, you are asked to find the absolute extrema between [-5,5] you see where the lowest points and highest points are on the graph. Most people will instinctively look at the large curves on the graph and choose the lowest point/highest point from them. These are the local maximum and local minimum for the entire graph but it is important to know that they will not always be the Absolute Max or Absolute Min. Why? Here is an example of what would make the previous statement true, Find the local extrema between [- \infty , \infty ] Would you say it those local max and min again? That would be incorrect. Because when you are faced with an infinite interval you must consider that the rest of the function goes on forever even if you do not see it on the graph there will never be an absolute maximum or minimum so long as you have an infinite interval to work with. The only time you will have an absolute extrema is when you are given a defined interval that you can work with and see where it is at its highest and lowest points.

 

 

Almost Lost It On This Kind Of Problem…

I’ve been trying to find out what I was doing wrong on the combined rules problems that involve the chain rule and the quotient rule. Hopefully, this blog will help prevent the stress I went through to figure out what I did wrong.

y= ( \frac {x+6}{x-2} )^5

First, you must use the quotient rule to set up the problem…

u= x+6

u'= 1

v = x-2

v' = 1

Now we find the derivative of the problem as a whole…

y= ( \frac {x+6}{x-2} )^5

y'= 5( \frac {x+6}{x-2} )^4

Now we multiply the derivative of the whole problem with the quotient rule we set up…

The finished quotient rule set up looks like…

\frac {dy}{dx} = \frac {-2-6}{(x-2)^2}

My big problem was that I was subtracting b from t and not the other way around. Not doing b(t')-t(b') will lead to an incorrect answer.

Put it all together…

\frac {dy}{dx} = 5( \frac {x+6}{x-2} )^4 \cdot \frac {-8}{(x-2)^2}

Now we multiply the two together to get the derivative.

= - \frac {40(x+6)^4}{(x-2)^6}

Remember, the denominator’s power is added by the power of the function being multiplied so x^4 \cdot x^2 = x^6

By Your Rules Combined…I AM CAPTAIN PLANET!!!

Dealing with problems that have multiple rules needed to solve takes quite a while to work out. Here, I hope to give an example to help anyone who has trouble understanding how to approach these problems.

Let’s use the problem…

y= x(x^5+7)^3

It doesn’t look that bad but one wrong move and the whole tower will come crashing down. Let’s start by determining what rules will be needed to solve this problem. Since it is two functions being multiplied, then we assume that the product rule is involved. However, x is being multiplied by a function that is an input for another function. That means we’ll have to use the chain rule to clean up that mess.

Now let’s start by using the product rule first…

u=x

u'=1

v= (x^5+7)^3

Now in order to find the derivative of u, we must use the chain rule…

v= (x^5+7)^3

With the chain rule you multiply the power ruled function by the derivative of what’s inside the function.

v'= 3(x^5+7)^2 \cdot 5x^4

= 15x^4(x^5+7)^2

Now we just plug the functions into the product rule…

\frac {dy}{dx} = x \cdot 15x^4(x^5+7)^2 + (x^5+7)^3 \cdot 1

= 15x^5(x^5+7)^2 + (x^5+7)^3

Now we factor out the (x^5+7)^2

= (x^5+7)^2[15x^5 + (x^5+7)]

Simplify…

= (x^5+7)^2(16x^5 +7)

 

Derivatives of Divide n’ Add Problems

Let’s see what happens when we have something added to our Quotient Rule problems.

y = \frac {9x-1}{4x+7} + x^6

Let’s find the numbers for the Quotient Rule

t = (9x-1)

t' = 9

b = (4x + 7)

b' = 4

So now let’s plug in the numbers…

\frac {dy}{dx} = \frac {(9x-1)(4) - (4x+7)(9)}{(4x+7)^2}

= \frac {36x + 63 - 36x -4}{(4x+7)^2}

= \frac {67}{(4x+7)^2}

Find the derivative of the function added to the fraction…

y= x^6

\frac {dy}{dx} = 6x^5

Now we add the derivative of the additional function…

\frac {dy}{dx} = \frac {67}{(4x+7)^2} + 6x^5