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Example 25

The perimeter of a rectangle is 100 cm. Find the dimensions of the rectangle when its area is maximum.

Solution : Let the length of the rectangle be 'x' cm and breadth be 'y' cm.

Let its area be A sq. cm.

Now given that perimeter = 100 cm

\ 2 ( x + y) = 100

\ x + y = 50

\ y = 50 - x

Then A= xy = x (50 - x)

By Differentiating w.r.to x, we get

= 50 - 2x

and = 0

gives 50 - 2x = 0

i.e. x = 25 (critical point)

also, f "(x) = -2

Þ f "(25) = -2 < 0

By the second derivative test, A is maximum at x=25 Þ y=25

\ when the area is maximum, the rectangle is a square of side 25 cm.


Example 26

Divide 15 into two parts such that the product of the square of one and the cube of other is maximum.

Solution :

By the second derivative test ‘f’ has a relative maximum when x = 9

Also, when x = 9

then y = 15 - 9

= 6

\ The two required parts of 15 are 9 and 6.

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Index

5.1 Tangent And Normal Lines
5.2 Angle Between Two Curves
5.3 Interpretation Of The Sign Of The Derivative
5.4 Locality Increasing Or Decreasing Functions 5.5 Critical Points
5.6 Turning Points
5.7 Extreme Value Theorem
5.8 The Mean-value Theorem
5.9 First Derivative Test For Local Extrema
5.10 Second Derivative Test For Local Extrema
5.11 Stationary Points
5.12 Concavity And Points Of Inflection
5.13 Rate Measure (distance, Velocity And Acceleration)
5.14 Related Rates
5.15 Differentials : Errors And Approximation

Chapter 6





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