Maxima of product of two log functions

Question

Is there a way to find the maxima of the product of two log functions. I need to find the maxima of:

$f(x) = log\frac{1000}{x+k}.log(x+1) \text{, where k is a constant < 500, and } x \in [0,1000] \text{ such that } x+k \leq 1000$

It is getting overly complex trying to set its derivative to 0.

graph of the function looks like this.

Edit: Pl see the updated constraint on x and k.

Answer 1

Let $c = \log_2 1000$. Then $f(x) = c\log_2(x+1) - \log_2(x+k)\log_2(x+1)$. Differentiating and equating to zero, we get $$ \frac{c}{x+1} - \frac{\log_2(x+k)}{x+1} - \frac{\log_2(x+1)}{x+k} = 0 \implies c(x+k) = \log_2[(x+1)^{x+1}(x+k)^{x+k}] $$ The RHS is a monotonically increasing, concave-up function. So, the graphs of $$y=c(x+k) \quad \text{ and } \quad y = \log_2[(x+1)^{x+1}(x+k)^{x+k}]$$ intersect at most once. We also have that $$\lim_{x \to 0^+} f(x) = 0 \quad \text{ and } \quad \lim_{x \to +\infty} f(x) = -\infty,$$ and that $f(1) > 0$ when $k$ is in the range you mentioned ($1 < k < 500$). Hence, there is at least one critical point of $f$. Thus, there is a unique critical point, which will be the point of maximum of the function. Hence, a maximum always exists.

Answer 2

You have $$f(x) = \log \left( \frac{1}{x+k} \right) \log \left( x+1 \right) = - \log(x+k)\log(x+1)$$

and $\log$ is an increasing function, so its maximum is attained for the smallest value of $x$, i.e. $x=1$, and its value is then $-\log(k+1) \log(2)$.