While testing a new car engine, engineers calculated that it's fuel ef

The standard way to solve these kinds of problems without using either calculus or specialized quadratic formulas (neither of which you need on the GMAT) is by "completing the square". I don't think I've ever needed to do that on a real GMAT problem, and the numbers in this question make it slightly awkward here, but we can. We know fuel efficiency is 97s - s^2. We can rewrite this by adding a number so we get something matching the (x - y)^2 = x^2 - 2xy + y^2 pattern:

\(\\
97s - s^2 = -(s^2 - 97s) = -(s^2 - (2)(48.5)s) = - (s^2 - (2)(48.5)s + 48.5^2 - 48.5^2) = -(s^2 - (2)(48.5)s + 48.5^2) + 48.5^2 = -(s - 48.5)^2 + 48.5^2\\
\)

We now have a square, (s - 48.5)^2, which has a minimum possible value of zero, so when we make it negative, to get -(s - 48.5)^2, that will have a maximum value of zero, precisely when s = 48.5. So we get maximum fuel efficiency when s = 48.5, and that maximum value is equal to 48.5^2.

But as I said above, I can't recall ever using this technique on a real GMAT problem (and I've definitely never used either of the techniques in the other two perfect solutions above), and if it were ever needed, you certainly wouldn't end up with numbers like "48.5" in the answer, so this isn't a realistic problem.

Given: While testing a new car engine, engineers calculated that it's fuel efficiency increased with speed up to a certain point and then decreased and that the efficiency was equal to 97 times the speed minus the square of the speed.

Asked: At what speed will the efficiency be maximum?

\(E = 97s - s^2 \);
\(\frac{dE}{ds }= 97 - 2s = 0 \)
\(s = \frac{97}{2} = 48.5 \)

IMO E

There is no need to use calculus or quadratic formulas to solve this problem. The value of an expression of the form (ax - x^2) where 'x' is the variable and 'a' is the coefficient is maximized when x=a/2. It is pretty simple once it is determined that the efficiency is equal to (97s - s^2)which is also a simple deduction.

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