23 Maths -- Derivatives and its Application

Find two numbers whose sum is 10 and the sum of whose squares is minimum.

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A gardener having 120 m. of fencing wishes to enclose a rectangular plot of land and also to erect a fence across the land parallel to two of the sides. Find the maximum area he can enclose.

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

Perimeter= x+x+y+y+x

120=3x+2y

y=60-3x/2

Let A be the area of the rectangle

A=x.y

A=x(60-3x/2)

A=60x-3x²/2

dA/dx=60-3x

d²A/dx²= -3<0 [Maximum]

Since

dA/dx=0

60-3x=0

x=20 cm

y=60 - 3/2 20 =30 cm

Now,

Area=x.y =2030 =600 cm²

Show that the rectangle of largest possible area, for a given perimeter, is a square.

Let X be the length and y be the breadth of rectangle and P be it's perimeter.

P=2(x+y)

P/2=x+y

y=P/2-x

If A be it's area then,

A=x.y

A=x.(P/2-x)

A=Px/2 - x²

dA/dx=P/2 - 2x

d²A/dx²= -2<0 [Maximum]

Since

dA/dx=0

P/2 - 2x=0

x=P/4

Also, y=P/2 - x =P/2-P/4 = P/4

Since length and breadth are equal, the rectangle is a square.

A closed cylindrical can is to be made so hat its volume is 52 cm3 . Find its dimensions if the surface is to be a minimum.

Let r be the radius, h be the height and V be the volume of the cylinder.

V=πr²h

52=πr² h

h=52/πr²

Let S be the surface area of cylinder

S=2πr(r+h)

S=2πr² + 2πrh

S=2πr² + 2πr * 52/πr²

S=2πr² + 104/r

dS/dr = 4πr-104/r²

d²S/dr² = 4π+208/r³ >0 (Minimum)

Since

dS/dr = 0

4πr-104/r2 =0

4πr³ -104=0

4πr³=104

r³=26/π

r=(26/π)^1/3

And,

h=52/πr²

h= 52/{π (26/πr)}.

h=2r = 2(26/π)^1/3

A man who has 144 metres of fencing material wishes to enclose a rectangular garden. Find the maximum area he can enclose.

Let x be the length and y be the breadth of a rectangular garden and P be it's perimeter.

P=2(x+y)

144=2(x+y)

72=x+y

y=72-x…......(1)

If A is its area then,

A=x.y

A=x(72-x)

A=72x-x²

dA/dx = 72-2x

d²A/dx² = -2 (Max)

Since,

dA/dx = 0

72-2x=0

x=36m

y=72-x =72-36 =36m

Area= x * y = 36*36 =1296 m²

A window is in the form of a rectangle surrounded by a semicircle. If the total perimeter is 9 metres, find the radius of the semicircle for the greatest window area.