x + y = 8 and P = 5x^2 + 11y^2, where x, y > 0. What is the minimum possible value of P?|
Yes ans is 220 as follows.
P=5x^2 + 11y^2
we will get
So P will have a minimum value.And that min value will be at x=(-b)/2a = 176/32 = 11/2
Putting x=11/2 we get y=5/2
put these values in P=5x^2 + 11y^2
We will get Pmin=220.
rahul, alternate method by differentitaion could be, p= 5x^2+11(8-x)^2
further dp/dx = 32x-176 =>dp/dx=0gives x=11/2
d2p/dx2 =32 which is positive, means minima exists at dp/dx,
put x=11/2 and get the value as 220.
you can also follow method od quadratic equations as mentioned by Ramesh, cheers ramesh!!