Problem 14

We first use the formula of the volume of a rectangular box.

V = L * W * H


The box to be made has the following dimensions:

L = 12 - x

W = 10 - 2x

H = x


We now write the volume of the box to ba made as follows:

V(x) = x (12 - 2x) (10 - 2x) = 4x (6 - x) (5 - x)

= 4x (x 2 -11 x + 30)


We now determine the domain of function V(x). All dimemsions of the box must be positive or zero, hence the conditions

x > = 0 and 6 - x > = 0 and 5 - x > = 0


Solve the above inequalities and find the intersection, hence the domain of function V(x)

0 < = x < = 5


Let us now find the first derivative of V(x) using its last expression.

dV / dx = 4 [ (x 2 -11 x + 3) + x (2x - 11) ]

= 3 x 2 -22 x + 30


Let us now find all values of x that makes dV / dx = 0 by solving the quadratic equation

3 x 2 -22 x + 30 = 0


Two values make dV / dx = 0: x = 5.52 and x = 1.81, rounded to one decimal place. x = 5.52 is outside the domain and is therefore rejected. Let us now examine the values of V(x) at x = 1.81 and the endpoints of the domain.

V(0) = 0 , v(5) = 0 and V(1.81) = 96.77 (rounded to two decimal places)


So V(x) is maximum for x = 1.81 inches. The graph of function V(x) is shown below and we can clearly see that there is a maximum very close to 1.8.

Problem 13

Which gives

| [cos a - cos b] / [a - b] | <= 1


But

| [cos a - cos b] / [a - b] | = |cos a - cos b| / |a - b|


When combined with the above gives

|cos a - cos b| / |a - b| <= 1


Multiply both sides by |a - b| to obtain


|cos a - cos b| <= |a - b|

Problem 12

Function cos x is continuous and differentiable for all real numbers. Use the mean value theorem, using 2 real numbers a and b to write

(cos x) ' = [cos a - cos b] / [a - b]


Take the absolute value of both sides

| (cos x) ' | = | [cos a - cos b] / [a - b] |


(cos x)' = - sin x, hence.

| (cos x) ' | < = 1

Problem 11

Use the mean value theorem to prove that for any two real numbers a and b,



| cos a - cos b| <= | a - b|

Problem 10

Let us now find f '(x).

f '(x) = -6x 2 + 6


We now construct an equation based on f '(c) = [f(b) - f(a)] / (b - a)

-6c 2 + 6 = -2


Solve for c to obtain 2 solutions

c = 2 sqrt(2/3) and c = - 2 sqrt(2/3)

Problem 9

f(x) is a polynomial function and is continuous and differentiable for all real numbers. Let us evalute f(x) at x = -2 and x = 2

f(-2) = -2(-2) 3 + 6(-2) - 2 = 2

f(2) = -2(2) 3 + 6(2) - 2 = - 6


Evaluate [f(b) - f(a)] / (b - a)

[f(b) - f(a)] / (b - a) = [ -6 - 2 ] / (2 - -2) = -2

Problem 8

If f is a function continuous on the interval [ a , b ] and differentiable on (a , b ), then at least one real number c exists in the interval (a , b) such that



f '(c) = [f(b) - f(a)] / (b - a).


The mean value theorem expresses the relatonship between the slope of the tangent to the curve at x = c and the slope of the secant to the curve through the points (a , f(a)) and (b , f(b)).

Problem 1:Find a value of c such that the conclusion of the mean value theorem is satisfied for



f(x) = -2x 3 + 6x - 2