Find two numbers whose sum is 10 and the sum of whose squares is minimum.
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.(P/2-x)
A=Px/2 - x2
dA/dx=P/2 - 2x
d2A/dx2= -2<0 [Maximum]
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)
dS/dr = 0
4πr-104/r2 =0
4πr3-104=0
4πr3=104
r3=26/π
r=(26/π)1/3
And,
h= 52/{π (26/πr)}.
h=2r = 2(26/π)1/3
Let x be the length and y be the breadth of a rectangular garden and P be it's perimeter.
144=2(x+y)
72=x+y
y=72-x…......(1)
If A is its area then,
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
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