Scribe Post 11/29

We started off class today with a quiz on IW's 1-5. After reviewing some quick questions on the quiz we moved into the lesson of the day and I was forced into being scribe for the day.

Finally we are using our knowledge of derivatives and applying it to more realistic problems. Word Problems, which Mr. O'Brien says "are really easy if you like word problems." He also promised the more we work on them the easier they will become. 

Optimization Questions- are questions which you find the max or min. 

ex/ If two numbers multiply to be 50 and If their sum is a min what are the numbers? What are the numbers if their sum is a max?

It is easiest to solve this problem by breaking it into two different equations:

1. Primary equation- let x be one number and let y be another. 

2. Secondary equation- relates a couple of variables using parameters. 

Solving the problem:

Rework the secondary equation into terms of y.

Substitute y into the primary equation and simplify.

Finding the derivative of the new equation and setting it to zero will help us find the CP's. setting equal to zero or zeros are

Finding where the derivative is undefined helps us find the other CP's  when x=0

By plugging in numbers to the derivative we can find the sign of the slope between the CP's 

There are many numbers which multiply to 50, yet only a couple have the potential to be min's or max's. Since when x=0 the function is undefined therefore   are potential answers

Since there is no smallest negative number which multiplies to 50 and there is no highest number which multiplies to 50:

If considered just negative numbers for the graph of S then the maximum values are the answer  or the two numbers -√50 and -√50

If considered just positive numbers for x then the minimum values are the answer

or the two numbers √50 and √50

ex/ The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a) Express the y coordinate of P in terms of x. b) Express the area of the triangle in terms of x. c) What is the largest area the rectangle can have, and what are its dimensions?

Not always but most of the time these situations can be put into primary and secondary equations. 

- The primary equation can be found using the diagram and the dimensions given by coordinates to find the area in terms of variables x and y plus the dimensions of the triangle.

- The secondary equation can be found using using the coordinates to find the slope of the side of the triangle.

By plugging in the secondary equation to the primary equation we can find the area in terms of x. and simplified

Since we are finding the largest area of the rectangle which fits in this triangle then we can find the largest area by finding the maximum. Using skills from the past we can find max where the derivative equals zero. set to zero zero is

We can then plug into the original equation to find y (the largest area). plug in x equals

Therefore to find the dimensions you can plug back into the equation. Since the base is 2x then and x=1/2 then the base is 1/2, therefore the height is 1.

ex/ A 20' high trough with 1' sides and a 1' bottom. The angle which controls width of the trough is the only things which can be varied. What value of the angle will maximize the trough's volume?

The equation can be given by the volume.

Substituting in for base and height we can find an equation in terms of the angle. using our knowledge of angles we can substitute

The derivative will help us find CP's  which simplifies to
Where the derivative of the volume changes from positive to negative is where there is maximum 



This can be found by setting the derivative to zero and using the pythagorean identities
simplify and set to zero equals set 2sinx-1 to zero
Maximum is Volume when

Tonight's homework is IW#6: p. 231/5, 9, 13, 17, 20, 22, 31, 41, 47, 53, 55, 56