Question

Maximum value of y in y=50x-5x^2

Answer 1

the minimum value of

y= 5x^2 -4x +3

If we take y=0 , then x becomes non real as 'D' or (b^2 -4ac) becomes < 0

If we take y = for any negative ( -ve )value , x becomes non real

=> y > 0 ,

y = 5x^2 -4x +3

Here, in the above function we need the minimum positive value of the function

=> 0 = 5x^2 -4x +(3-y)

x = {-b +,- √(b^2 - 4ac)} / 2a

=> x = { 4 +,- √(16 - 4*5*(3-y) ) } /10

If we make D= 0 , we get the minimum value of the function.

For making D= 0

4ac = 16

=> 20(3-y) = 16

=> 3-y = 16/20= 4/5

=> y = 3 - 4/5

=> y = 11/5

Minimum value of the function= 11/5

Answer 2

Answer:

Maximum value of y in y=50x-5x^2