WORD PROBLEMS ON APPLICATION OF DERIVATIVES CALCULUS

Problem 1 :

A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at other sides of the page is 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum.

Solution :

Given :

Area of the rectangle  =  24 cm²

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

Horizontal measure  =  x

Including margin  =  x+2

Vertical measure  =  y

Including margin  =  y+3

xy  =  24

y  =  24/x

Area of the paper  =  (x+2) (y+3)

f(x)  =  (x+2)((24/x)+3)

f(x)  =  30 + 3x+48/x

f'(x)  =  3-48/x²

f'(x)  =  0

48/x²  =  3

x²  =  16

x  =  4, -4

f''(x)  =  0+96/x³

f''(x)  =  96/x³

f''(-4)  =  96/(-4)³ < 0 (Maximum)

f''(4)  =  96/(4)³ > 0 Minimum

When x  =  4, y = 24/4

y  =  6

Length including margin  =  4+2  ==> 6

Width including margin  =  6+3  ==> 9

Length and width of the rectangle is 6 cm and 9 cm.

Problem 2 :

A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq.mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?

Solution :

Area  =  180000

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

xy  =  180000

y  =  180000/x

Perimeter of the fencing  =  2x + y

P(x)  =  2x+180000/x

p'(x)  =  2 - 180000/x²

p'(x)  =  0

180000/2  =  x²

90000  =  x²

x  =  300

p''(x)  =  -360000/x³

p''(300)  =  -360000/(300)³ < 0 (Maximum)

y  =  180000/300

y  =  600 m

Length of the material needed  =  2(300) + 600

=  600 + 600

=  1200 m

Problem 3 :

Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10 cm.

Solution :

PQ  =  2(10 cos a)  ==>  20 cos a

SP  =  2(10 sin a)  ==>  20 sin a

Area of PQRS  =  (20 cos a) (20 sin a)

A(a)  =  200(sin 2a)

A'(a)  =  400 cos 2a

A'(a)  =  0

400 cos 2a  =  0

2a  =  cos^-1(0)

2a  =  π/2

a  =  π/4

A''(a)  =  -800 sin 2a

A''(a)  =  -800 sin 2(π/4)

A''(a)  =  -800 < 0 (Maximum)

PQ  =  20 cos π/4  =  20/√2  =  10√2

SP  =  20 sin π/4  =  20/√2  =  10√2

So, the dimensions are 10√2 cm, 10√2 cm.

Problem 4 :

Prove that among all the rectangles of the given perimeter, the square has the maximum area.

Solution :

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

Perimeter of the rectangle  P  =  2(x+y)

x+y  =  p/2

y  =  p/2 - x

Area of the rectangle f(x)  =  x y

=  x(p/2 - x)

=  xp/2 - x²

f'(x)  =  p/2 - 2x

f'(x)  =  0

p/2 - 2x  =  0

2x  =  p/2

x  =  p/4

f''(x)  =  0 - 2 < 0 (Maximum)

y  =  p/2 - x

Applying the value of x in the equation above, we get

y  =  p/2 - p/4

y  =  p/4

So, it is square.

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