APPLICATION OF DIFFERENTIATION MAXIMUM AND 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/x²  =  0

20/x²  =  1

x²  =  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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