Maximum power point, why do you need to change voltage again? Isn't it simply \$ P = i^2 \cdot R \$?
Here's a realistic curve for a realistic solar "panel" that you can get from a variety of hobbyist suppliers. This particular one has 9 solar cells (a 3x3 grid) wired up in series.



Now, the blue curve represents the current vs voltage for the device. But what does it really mean? Well, assuming full sunlight (whatever that means for this curve) is hitting it, then:


If you short-circuit the output using an ammeter to do it, then there will be about \$80\:\text{mA}\$ current as measured with an ammeter across the solar cell terminals. Note here that, by definition, the output is shorted so the voltage at the output is going to be about \$0\:\text{V}\$. If you look at the blue curve, this is the left-most end of it. The power delivered by the solar cell into the ammeter will be close to \$0\:\text{W}\$ because although there is current flowing, there is no voltage across the ammeter. So the power delivered to the ammeter is close to zero.
If you leave it open-circuit and measure the voltage with a high impedance voltmeter, then you will measure about \$4.2\:\text{V}\$. Note here that, by definition, the output is essentially open, so the current at the output is going to be about \$0\:\text{mA}\$. If you look at the blue curve, this is the right-most end of it. The power delivered by the solar cell into the voltmeter will be close to \$0\:\text{W}\$ because although there is voltage present, there is no current drawn through the voltmeter. So the power delivered to the voltmeter is close to zero.


Now, ammeters and voltmeters are ideal devices for the work they do. But they aren't designed or intended to dissipate power from circuits. They are supposed to do their work with the least possible disturbance to it. So they do a job very well. But not the job you want, here.

Now, look at the curve at about where the blue curve crosses through about \$2.7\:\text{V}\$. At that point, the curve also shows about \$60\:\text{mA}\$. Clearly, the load here must be \${{2.7\:\text{V}} \over {60\:\text{mA}}}=45\:\Omega\$. So that must have been the resistor used there to measure that data point for the curve. But this also means \$\left(60\:\text{mA}\right)^2\cdot 45\:\Omega=162\:\text{mW}\$! That's more than zero power into the load!

If you try out different resistors (loads) across the solar cell shown above, you will get different values of both voltage and current. For example, if you tried out an \$82\:\Omega\$ resistor, you'd probably get \$3.3\:\text{V}\$ and \$40\:\text{mA}\$. But this would only be \$132\:\text{mW}\$.

By tracing your fingers over each point along the blue curve and dividing the voltage at a point by the current also at that point, you can work out what resistor was used there. And you can work out the power. Or you can just multiply the voltage by the current for each point on the blue curve. If you do that, you will get the red curve. This is the power that would be delivered into the appropriate resistor which allows the solar cell to develop a certain voltage and current with that resistor value.

However, the curve isn't the same for any two solar cells. Nor is it the same with different temperatures. Nor is it the same with different light striking it. The blue curve varies with many different situations. An MPP circuit attempts to trace out the curve for you, in each and every one of these special cases (solar cell, temperature, lighting, etc) and searches for and holds near to the maximum power point -- the highest part of the red curve. Unfortunately, there is no specific voltage nor any specific current that always works. So the circuit has to constantly "hunt" around in order to stay close to it.