x greater than zero and y greater than zero x+y=32 and x*2 +y*2 is minimum then x,y is
Answers
Answer:
x=16,y=16
(I have provided two answers, in case your not familiar with calculus jump to alternate method)
Step-by-step explanation:
Given x+y =32 ⇒y=32-x, and x>0 and y>0
Required minimum value of
x²+y² let it be s
x²+(32-x)²=s
for finding minimum differentiate the above expression with respect to x
2x-2(32-x)=0
4x=64
x=16
and also y=16 (x+y=32)...........
If you are not familiar with calculus use the below procedure,
ALTERNATE METHOD
x²+y²=s
(x+y)²-2xy=s ( since (x+y)²=x²+y²+2xy)
32²-2xy=s
2xy=32²-s..............(1)
use AM≥GM (for x,y)
x+y/2≥√xy
16≥√xy (x+y=32)
16²≥xy
( observe x²+y²=(x+y)²-2xy, x²+y² has minimum value when 2xy has maximum value)
for x²+y² to be minimum
16²=xy
from (1)
2 x 16²=32²-s
s=16²(2²-2)
s=16²(2)
i.e x²+y²=16²(2)
x²+(32-x)²=16²(2)
x²+32²+x²-64x=16²(2)
2x²-64x+16²(4-2)=0
2x²-64x+16²(2)=0
x²-32x+16²=0
It is quadratic equation,
x²-16x-16x+16²=0
x(x-16)-16(x-16)=0
(x-16)(x-16)=0
therefore x=16 and also since x+y=32
y=16