There is some discussion in the comments above about another way to think of the scenario in Example 2.

Think of the parabola as a curved wall in a building, and imagine there is an electrical pole at position (1,4). The question is basically saying, if we want to use the least amount of wire, and need to run a wire from the electrical pole to the building, where should the wire go?

]]>If so, here is the detail, which basically involves playing around with exponents:

r = (500/pi)^{1/3}

r^2 = ((500/pi)^{1/3})^2

= (500/pi)^{2/3}

= 500^{2/3} / pi^{2/3}

Substitute into h = 1000/(pi*r^2) to get

h = 1000 / (pi * (500^{2/3}/pi^{2/3}))

= 2*500 / pi^{1/3}*500^{2/3}

= 2*500^{1/3} / pi^{1/3}

= 2*(500/pi)^{1/3}

= 2*cuberoot(500/pi)

= 2*r.

Multiple critical values are possible, although don’t generally occur in the types of problems we do in section 4.7.

]]>Example 3 uses the distance formula. To find the distance between two points, find the distances between the x- and y-coordinates, square them, and add them; then take the square root.

In general, distance between (x1,y1) and (x2,y2) is

d = sqrt((x2-x1)^2 + (y2-y1)^2)

This is really just the pythagorean theorem if you draw a picture. d is the hypotenuse of a right triangle with base x2-x1 and height y2-y1.

]]>The smallest possible value of x is 0 (this means running all 8km in the picture) and the largest possible value of x is 8 (this means rowing all the way, and not running at all). From section 4.1, the abs. minimum time occurs at a critical number (x=9/sqrt(7)) or at an endpoint (x=0,8). At the bottom of page 329, they plug these three values in T(x) to see which one produces the minimum time.

]]>Generally, the endpoints in these types of problems produce unrealistic situations (but are “theoretically possible”). So for the most part, you are safe with assuming that the number you’re looking for is the critical number.

]]>For example, if the minimum of d is 2, then the minimum of d^2 is 4. If you take the input that gives you d=2, then that same input gives you d^2=4.

The”First Derivative Test for Absolute Extreme Values” is just saying that if there is only one “change” (either inc/dec or dec/inc), then the max/min must be absolute.

For example, if a function is increasing and then decreasing (and that’s the only change), then the place where the change happens has to be the highest point (abs. max).

If a function is decreasing and then increasing (and that’s the only change), then the place where the change happens has to be the lowest point (abs. min).

Generally, you can think of it like this. If there is only one critical number, then local max = abs. max and local min = abs. min.

]]>