## Friday, April 27, 2012

### Calculus Revisited #7: Finding Extrema

We have arrived at the 7th in a 21 blog entry in our Calculus Revisited series.

Today - we will use the derivative to find local extrema. Extrema include the maximum and minimum values of a function f(x). Finding extrema determines where the function f(x) "turns".

First Derivative Test: A process for finding extreme of the function f(x):

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Compute f'(x) where x is near c but not at x = c.

(A) If f'(x) < 0 when x < c and f'(x) > 0 when x > c, then f(x) has a minimum at x = c.

(B) If f'(x) > 0 when x < c and f'(x) < 0 when x > c, then f(x) has a maximum at x = c.

(C) If (A) or (B) do not apply, then f(x) has an inflection point at x = c.

Second Derivative Test: Another process, this one involving the second derivative:

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Calculate f''(c) (second derivative).

(A) If f''(c) >0, then f(x) has a strict local maximum.

(B) If f''(c)<0, then f(x) has a strict local minimum.

(C) If (A) or (B) do not apply, the second derivative test is inconclusive.

Some observations:

1. If f'(x) > 0, f(x) is said to be increasing at x.
2. If f'(x) < 0, f(x) is said to be decreasing at x.
3. If f''(x) > 0, f(x) is said to be concave up at x. (think of a bowl holding water)
4. If f''(x) < 0, f(x) is said to be concave down at x. (think of a bowl turned upside down)

Problems
1. Find extrema of the function y(x) = -x^3 + 2x + 1

Find the critical points
y'(x) = 0
d/dx (-x^3 + 2x + 1) = 0
-3x^2 + 2 = 0
-3x^2 = -2
x^2 = 2/3
x = ±√(2/3)

The critical points are x = √(2/3) and x = -√(2/3).

By the second derivative test:
y''(x) = -6x

y''(√(2/3)) = -6 * √(2/3) < 0. f(x) has a maximum at x = √(2/3).

y''(-√(2/3)) = -6 * -√(2/3) = 6 * √(2/3) > 0. f(x) has a minimum at x = √(2/3).

√(2/3) ≈ 0.81650

2. An open box will be made out of a 12" x 10" sheet of cardboard. (see drawing below, not drawn to scale). A square of length x inches will be cut from each corner of the sheet. What should be the dimensions to construct a box with maximum volume?

x is going to be restricted. First of all x > 0. Second, x < 5. (because 10 - 2 * 5 = 0).

The volume of the box is:
V = (12 - (x + x)) * (10 - (x + x)) * x
= (12 - x^2) * (10 - 2x) * x
= 4x^3 - 44x^2 + 120x

To find the maximum volume, find the critical points. Use the derivative of V in terms of x.

dV/dx = 0
12x^2 - 88x + 120 = 0
x = (88 ± √(88^2 - 4*12*120))/(2*12)
where:
x ≈ 1.8107 and x ≈ 5.5226.

The latter measurement does not make sense since 10 - 2*5.5226 < 0, and negative dimensions do not make sense.

Test point x ≈ 1.8107.

The second derivative:
d^2V/dx^2 = 24x- 88

Then 24(1.8107) - 88 < 0. The test point is a local maximum.

The dimensions of this box are approximately 8.3786'' x 6.3786'' x 1.8107'' with a volume of about 96.7706 in^3.

3. Minimize the distance between the point (3,1) and the line y(x) = 2x + 5. (see diagram below).

The distance formula between two points (x1, y1) and (x2, y2).

D = (x2 - x1)^2 + (y2 - y1)^2

Find the distance between (3,1) and (x, 2x+5).

D = (x - 3)^2 + (2x + 5 - 1)^2
= (x - 3)^2 + (2x + 2)^2
= 5x^2 + 2x + 13

Taking the derivative D'(x)

D'(x) = 0
10x + 2 = 0
10x = -2
x = -1/5

Since D''(x) = 10 > 0, the point x is a strict local minimum.

The minimum distance is 5 * (-1/5)^2 + 2 * (-1/5) + 13 = 68/5 = 13.6

Next time we deal with problems involving related rates. Until next time, Eddie.

This blog is property of Edward Shore. © 2012

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