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. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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