Maximize y, minimize x on logarithmic growth curve Sorry for the noob question. This is my first question here. It's been years since calc and since I'm not in a course right now, I don't really have any insight on how to solve this as there's no "last chapter focused on that, and on this chapter, we're focusing on this" thing I can go off of. This is n econ question (not a graph of profit over cost) this is just an example as the graph has a similar curvature.
Let's say there is a curve where y (something like profit) increases as x (something like time invested or the cost) increases. As x increases, the amount that y increases becomes smaller and smaller.
I want to find the point where maximum profit can be achieved while avoiding a low return on investment (at some point in the +x direction, it becomes a waste of resources.)
I feel like I need another constraint. A friend suggested building a linear regression line and finding the points of intersection, but given the nature of the curve, I feel linear regression is not appropriate here.
PS: This is not homework. It's just a question I've had in my head. I made it through calc I, II, and multivariate years ago so I feel like an idiot for not being able to unfurl this. I've taken a look at the derivatives of similar curves (I don't have an equation for an increasing curve (but barely increasing as x increases) to really work with. A quick look at the graphs of derivatives didn't offer any insight. There is no inflection point.
My gut says that for this particular image I found, the point near (20,40) looks promising, but I can't flesh out why.
See this graph as an example:

 A: This is more of an Econ problem than a calc one. The only thing from calc that would help was the idea of marginal profit aka rates aka the first derivative. So referring to your question, if you think about it, of course, you'll make more profit the more you invest which is what the original graph represents. So the real profit-maximizing choice is whatever your max budget is since either way you'll be making more money the more you invest. Rather, like you suspected, in real-life, you'll have more variables than just cost vs profit.
Consider comparing that graph to another graph of a different investment. Now you need to know what amount to invest in each to make the most money. This can be solved.

(^Replace "utility" with "profit." The calculations are the same.)
Or consider having two graphs. One is quantity (inventory, workers, factories, etc) vs marginal revenue (revenue, not profit, per unit of quantity), and the other should be quantity vs marginal cost. The Profit-Maximizing Quantity is the quantity where $MR = MC$.

Or consider this from an econ textbook:

"First consider the upper zone, where prices are above the level where marginal cost (MC) crosses average cost (AC) at the zero-profit point. At any price above that level, the firm will earn profits in the short run. If the price falls exactly on the break-even point where the MC and AC curves cross, then the firm earns zero profits. If a price falls into the zone between the break-even point, where MC crosses AC, and the shutdown point, where MC crosses AVC, the firm will be making losses in the short run—but since the firm is more than covering its variable costs, the losses are smaller than if the firm shut down immediately. Finally, consider a price at or below the shutdown point where MC crosses AVC. At any price like this one, the firm will shut down immediately, because it cannot even cover its variable costs."
A: If we interpret your graph as "if I put in $x$ buckos, I will recieve $y=f(x)$ buckos", then you should put in more $x$ if $f(x)-x$ is increasing at $x$.  I have drawn the line $y=x$ in red (as the scales used to draw the $x$ and $y$ axes are not the same, the $y=x$ line is not at 45 degrees):

the vertical distance between points on the the green and red curves seems to indeed be maximal at about (20,40). Since $f$ is a nice smooth function, the condition that $f(x)-x$ is increasing is the same as the condition that its derivative is positive, i.e. $f'(x)>1$. Since $f$ is increasing, the best point is when $f'(x)=1$, and then from that point onwards, its a waste of resources. We can find this point graphically -

So it turns out, you should be using $x$ a little bit smaller than $20$.
