A man who has 144 metres of fencing material wishes to enclose a rectangular garden. 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

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.

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

V=πr² h

52=πr²h

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

Let x and y be the numbers.

x+y=10

y=10-x

Let S be the sum.

S= x²+ y²

S= x²+ (10-x)²

dS/dx = 2x+2(10-x) *(-1)

dS/dx = 2x-20-2x

dS/dx = 4x-20

d²S/dx²= 4>0 (Minimum)

dS/dx =0

4x-20=0

4x=0

x=5

y=10-x = 10-5 =5

Hence the two numbers are 5 and 5