The power (P) delivered to a load by a current of i amperes through a resistance of 9 ohms (when 110V are supplied at one end of the line) is given by the formula P= 110i -9i^2.

Question

Determine the maximum power that can be delivered. Give your answer to four decimal places.

Answer 1

Hi Seth,

Points of local extrema (maximum or minimum) are found by letting the FIRST derivative zero. After finding such a point, calculate the SECOND derivative at that point. If the second derivative turns out NEGATIVE, then at that point the function reaches local MAXIMUM. In the opposite case OF negative second derivative we would be dealing with a local MINIMUM.

So, find the point of zero 1st derivative : (we'd use just values for i but keep in mind when reporting results that all numbers mean hear number of amps (A).

d/di(110i-9i^2)=110-9x2i=110-18i=0 giving us i=(110/18)=55/9(A)

Now, Is that a min or max?

Therefore, find 2nd derivative

d²/di²(110i-9i^2)=d/di(d/di(110i-9i^2))=d/di(110-18i)=-18<0

Therefore, the point i=55/9A is a point of maximum power

Now, the maximum power is found by using the current 55/9A in the expression for power

P_max=(110i-9i^2)|_i=55/9=110x55/9-9(55/9)^2=336.1(J)

(Note: Not clear why the 4th decimal place must be listed but the data are given with no indication of accuracy (strictly speaking, 9 Ohms means only 1 sig.fig. so the correct result

must have been 3x10^2 J but maybe your instructor had something else in mind)

Hope it would help

Best,

Dr.Ariel B.