Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.

The minimum value of px+qy when xy= r^2 is

The minimum value of (px+qy) when xy=n^(2) is equal to .

Find the maximum and the minimum values, if any, of the following functions, without using the derivative: x + 1/x , x > 0

Let f(x)=x^(2)+2px+2q^(2) and g(x)=-x^(2)-2px+p^(2) (where q ne0 ). If x in R and the minimum value of f(x) is equal to the maximum value of g(x) , then the value of (p^(2))/(q^(2)) is equal to

Given that q^2-p r ,0,\ p >0, then the value of |p q p x+q y q r q x+r y p x+q y q x+r y0| is- a. zero b. positive c. negative \ d. q^2+p r

Given that q^2-p r 0, then the value of {:|(p, q, px+qy), (q, r, qx+ry), (px+qy, qx+ry, 0) |:} is- a. zero b. positive c. negative \ d. q^2+p r

The lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then

The lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then