Highest value of equation "T(C-NT)"

Question

Given equation T(C-NT). Here given value of C & N. I have to find the minimum value of T so that given equation value is maximazied.

My approach:

Let maximum value of equation is y. So, y = T(C-NT) y = T C - T^2 N

If we differentiate this equation with respect to T then we got 0 = C - 2NT.

So, we can write T = C/(2N).

But I'm getting verdict Wrong Answer.

It is math question, but i implement it with C++. So, I thought it's related to C++. Should I remove C++ tag? — Abu Hanifa, Oct 11 '14 at 13:17

score 1 · Accepted Answer · answered Oct 20 '14 at 21:57

Let highest value of the equation is x. So, x = T(C-NT) If we simplify this equation x = TC-NT^2 Which is simply a inverted U curve. We have to find the highest value of T at which point the gradient equal to zero.

If we differentiate this equation with respect to T then, 0 = C-2NT. So, T = C/(2N)

But this value of T isn't correct answer. This value of T says optimal value of T must be equal or greater than this value.

So, we have to increment this value by 1 and check if we can get greater value of the equation.

Highest value of equation "T(C-NT)"

1 Answers1