APPLICATION OF DIFFERENTIATION MAXIMUM AND MINIMUM

Optimization is a process of finding an extreme value (either maximum or minimum) under certain conditions.

A procedure for solving for an extremum or optimization problems.

Step 1 :

Draw an appropriate figure and label the quantities relevant to the problem.

Step 2 :

Find a expression for the quantity to be maximized or minimized.

Step 3 :

Using the given conditions of the problem, the quantity to be extremized .

Step 4 :

Determine the interval of possible values for this variable from the conditions given in the problem.

Step 5 :

Using the techniques of extremum (absolute extremum, first derivative test or second derivative test) obtain the maximum or minimum.

Problem 1 :

Find two positive numbers whose sum is 12 and their product is maximum.

Solution :

Let one number be x. The other number is 12-x.

Let f(x) be the function

f(x)  =  Product of two numbers

f(x)  =  x(12-x)

f(x)  =  12x-x2

f'(x)  =  12-2x

f'(x)  =  0

12-2x  =  0

x  =  6

f''(x)  =  -2 < 0 Maximum

The first number is 6 and the another number is 6.

Product of two numbers  =  36.

Problem 2 :

Find two positive numbers whose product is 20 and their sum is minimum.

Solution :

Let x be the one number, then the another number  =  20/x

Let f(x) be the function.

f(x)  =  x + (20/x)

f'(x)  =  1 - 20/x2

f'(x)  =  0

1 - 20/x=  0

20/x=  1

x =  20

x  =  2√5

f''(x)  =  -40/x3 < 0

f''(2√5)  =  -1/√5 < 0 (maximum)

Another number  =  20/2√5   =  2√5

So, the sum of numbers  =  2√5 + 2√5  ==>  4√5.

Problem 3 :

Find the smallest possible value of

x2 + y2

given that x + y =10.

Solution :

Let f(x) be the function.

Given :

x+y  =  10

y  =  10-x

f(x)  =  x2 + y2

f(x)  =  x2 + (10-x)2

f(x)  =  x2 + 100+x2-20x

f(x)  =  2x2-20x+100

f'(x)  =  4x-20

f'(x)  =  0

4x-20  =  0

x  =  5

f''(x)  =  4(1) ==>  4  > 0 (Minimum)

y  =  10-x

y  =  10-5

y  =  5

x2+y2  =  52 + 52

x2+y2  =  50

So, the value of x2+y2 is 50.

Problem 4 :

A garden is to be laid out in a rectangular area and protected by wire fence. What is the largest possible area of the fenced garden with 40 meters of wire.

Solution :

Let x and y be the length and width of the rectangle.

2(x+y)  =  40

2x+2y  =  40

2y  =  40-2x

y  =  20 - x

Area of rectangle  =  x(20-x)

A(x)  =  20x-x2

A'(x)  =  20-2x

A'(x)  =  20-2x  =  0

x  =  10

A''(x)  =  -2 < 0 (Maximum)

y  =  20-10

y  =  10

Area  =  10(10)

=  100 m2

So, area of the rectangle is 100 m2.

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