Aim: More optimization problems

Do Now:
Walking Speed = 5 mph
Swimming Speed = 2 mph
How much time will it take to walk to point X and swim to point Y?
*Note: The diagram consists of two right triangles. The one with point X at the end of the hypotenuse has legs of length 2 miles and 1 mile. The one with point Y has legs of length 2 miles and 3 miles.
To find the time taken to walk to point X, use the Pythagorean Theorem.
The length of the hypotenuse is radical 5.
Using the same process the distance to Y is radical 13.
To find the total time needed, divide (radiacal 5)/5 + (radical 13)/2. This gives us 2.25 hours which is the answer.

We looked at three types of optimization problems:

Problem 1:
For a graph of y = sqrt(x), find the point on the curve that is closest to the point (1,0).
Any point on the curve of this graph should be (x, sqrt(x) )
To solve, this we can use the two coordinate points we know and use them in the distance formula.
external image ch1not28.gif
We can square both sides to get d^2 = (x2-x1)^2 + (y2-y1)^2.
If we plug in the points (x, sqrt(x) and (1,0), we get:
d^2 = (x-1)^2 + (sqrt(x)-0)^2
d^2 = x^2 - x + 1
Find the vertex using -b/2a, which gives us an x value of 1/2.
Therefore, y = sqrt(2)/2
Answer: ( 1/2, sqrt(2)/2)

Problem 2:
There is a piece of square cardboard, which is 20 in. by 20 in. We can create a box using this square by cutting off four equal squares from each corner with side length equal to x. What value of x will give us a box with the greatest volume?
V = lwh
The height is x. The length and width are equal so they are both 20-2x.
Therefore, the volume is:
V = (x)(20-2x)(20-2x)
V = 4x^3 - 80x^2 + 400x
Using the formula to find the turning points of a cubic curve:
we get x = 10 and x = 3 1/3
Using a graphing calculator, we can find the maximum which turns out to be 592.6

Problem 3:
The third problem was similar to the Do Now problem.
If walking takes 5 mph and swimming takes 2 mph, find the shortest amount of time required to get from point Z to point Y.
To find the walking time, start with the Pythagorean Theorem. The two sides of the right triangle are 2 and x. Therefore the hypotenuse is sqrt(4 + x^2).
For the swimming, the two legs are 3 and 3-x (since the entire base was 3 miles long). The hypotenuse is sqrt(x^2-6x+18).
To find the minimum time required, do:
sqrt(4+x^2) + sqrt(x^2-6x+18) = T
5 2
Using a graphing calculator to plot the minimum, we get 2.1 hours.

Manisha Basak
Period 10