Suppose that you have a sequence of purchases. Prove that the optimum time to buy each element in the sequence is as soon as you can afford it.
With that lemma in the bank, you can consider a set of states where a state is a tuple of the number of cursors, grannies, farms, cursor upgrades, and granny upgrades purchased. Form a digraph where each state has edges to the state with one more cursor, the state with one more granny, etc. Therefore each state has up to five edges. Add a new vertex "Target" and add an edge from each state to the target. Give the edges weights as the cost of the upgrade divided by the production rate of the source state.
Then you just need to do a standard shortest path search (Dijkstra's algorithm, A*, etc) in the graph. This should be much much much more efficient than the brute force you describe.
Here's a neat feature I didn't realize existed; you can ask WolframAlpha if it thinks a constant has a closed form and it will spit out suggestions. For your $0.8284 \dots$ it suggests
$$2 \sqrt{2} - 2 = \frac{2}{\sqrt{2} + 1} \approx 0.82842712474619 \dots $$
which is an exact match to all of your digits.
I don't understand your problem statement (I put a long list of questions into the comments) but fortunately the mathematical content can be reverse-engineered from the spreadsheet screenshots alone, together with the big hint that it should all come down to solving a quadratic equation. It looks like if you pay out $x$ credits for the third lemon then the spreadsheet reads
$$\begin{array}{cccc}
\text{assets} & \text{units} & \text{backing} & \text{cost} \\
\text{lemons} & 2 & 2x & x \\
\text{sugar} & 1 & 2 - x & 2 - x \\
\text{total} & 3 & 2 + x & 2 \\
\end{array}$$
(I don't understand this, especially the backing calculations) and the condition you want appears to be that the fraction of the cost (I don't understand what this means) made up by the lemons, which is $\frac{x}{2}$, is equal to the fraction of the backing (I don't understand why you're calculating this the way you are) made up by the sugar, which is $\frac{2 - x}{2 + x}$, so
$$\frac{x}{2} = \frac{2 - x}{2 + x}.$$
Clearing denominators gives a quadratic equation
$$x^2 + 4x - 4 = 0$$
which has unique positive real solution $x = 2 \sqrt{2} - 2$. But given the lack of precision in the problem statement I am not at all confident that this is the number you actually wanted.
Edit: Here is my current understanding of how this calculation generalizes to arbitrary numbers. Suppose we have $L$ lemons, $S$ ounces of sugar, and $C$ credits in circulation, and Mom wants to deposit $L'$ new lemons and we want to determine how many credits $x$ to give her. If I've understood everything correctly, the spreadsheet should now read
$$\begin{array}{cccc}
\text{assets} & \text{units} & \text{backing} & \text{cost} \\
\text{lemons} & L + L' & (L + L')x & x \\
\text{sugar} & S & C + (1 - L - L')x & \frac{C + (1 - L - L')x}{S} \\
\text{total} & L + L' + S & C + x & \frac{C + (1 - L - L' + S)x}{S} \\
\end{array}$$
(this shouldn't be telling you anything you didn't already know because you must already have formulas equivalent to these in your spreadsheet if I've correctly understood the problem). The fraction of the cost made up by lemons is now $\frac{Sx}{C + (1 - L - L' + S)x}$, while the fraction of the backing made up by sugar is now $\frac{C + (1 - L - L')x}{C + x}$, and the condition that these are equal is
$$\frac{Sx}{C + (1 - L - L' + S)x} = \frac{C + (1 - L - L')x}{C + x}$$
which is again, after clearing denominators, a quadratic equation in $x$.
Best Answer
First of all, scale down the various ingredient requirements from a 48-cookie batch size to an individual cookie size. This is done by dividing your three amounts {1.5, 1, 2.75} by 48, giving {1/32, 1/48, 11/192}. Then all one has to do is multiply the required cookie batch size by each of these individual amounts. At least that is how I'd do it.