Max possible probability of exponential distribution for given $\lambda$?

Question

The maximum possible probability an exponential distribution with an arbitrary parameter can give to the interval $[3,6]$ is :

$(a) \ \ \dfrac{1}{2}$

$(b) \ \ \dfrac{1}{3}$

$(c) \ \ \dfrac{1}{4}$

$(d) \ \ \dfrac{2}{3}$

My input: $X \sim exp(\lambda)$

$f_X(x)=\lambda e^{-\lambda x} ; x\ge0$

$F(6)-F(3)=e^{-3 \lambda}-e^{-6 \lambda}$

I started by putting different values of lambda and calculating probability. Which is very time consuming for the exam I am preparing for.

If I take $\lambda=\dfrac{1}{3}$ I get $0.232$

But there must be some other way to figure this out which I am missing.

Answer 1

Hint: You're trying to maximize the quantity $$h(\lambda):=e^{-3\lambda}-e^{-6\lambda}$$ over all $\lambda>0$. This can be solved with calculus: take the derivative, set to zero, check you have a maximum.