Derivative-free Approach

Consider equation (1)

W = 0.03t2 - 0.39t + 7.3 (1)

i.e.

W = 0.03(t2 - 13t + 730/3) (5)

Now, let us complete the squares on the right hand side of equation (5).

Then,

W = 0.03(t2 -13t + 132/22 - 132/22 + 730/3)

or

W = 0.03(t2 - 13t + 132/22) + 0.03(730/3 - 132/22)

i.e.

W = 0.03(t-13/2)2 + 0.03(730/3 - 132/22). (6)

If we look at the right hand side of equation (6), the first term 0.03(t-13/2)2 is either zero or positive for any t value and the second term is positive (check this) all the time. This means that W will have a minimum when the first term is zero (are you clear on this?)

i.e. W will have a minimum when

0.03(t - 13/2)2 = 0

or when

t = 13/2 = 6.5

and this is exactly the same answer we got earlier, in equation (4), by using the derivative approach!