Right. So in this problem we have the picture of a part of a parabola and it is defined by f of X equals four minus x squared. And then we're fitting a rectangle inside the parabola and it's going to have boundaries of the X axis and the Y axis. And the location where the one vertex of the rectangle meets that upside down parabola is called X. Y. The point down here is X zero. The point here is 00 And the point right here is going to be zero. Why? And the first thing we're going to do is we're going to write the perimeter of the rectangle as a function of X. We're then going to calculate the absolute minimum and maximum of that new function and then we're going to determine the dimensions of the rectangle with the largest perimeter. We're then gonna start all over and we're going to express the area as a function of X. Find the absolute minimum and maximum of that new function and then find the dimensions of the rectangle that yield the largest area. So let's think about what the definition of perimeter is. Definition of perimeter is to go all the way around the rectangle. So if I was to draw the rectangle and I knew from here to here we went X. That means our um base or length of this rectangle is X. And I know we went from here to here to find the height of that rectangle. So we're going to call it why but why is actually defined right here? Why is the same thing as for minus X squared? So therefore for part A we could say the perimeter of this rectangle is X plus four minus X squared plus X plus four minus X squared. Because we're going across up across and down to find our perimeter. So the perimeter, when I combine all my late terms would be negative two X squared plus two X plus eight. So that's part a part B is asking you to determine the absolute minimum and absolute maximum of that function on the closed interval from zero 22 And that's going to lead us to the extreme value theorem. And the extreme value theorem says that a continuous function must obtain absolute minimum and absolute maximum values on a closed interval. Yeah, it also says these extreme values. Mhm. Yeah. Are obtained either on a relative extra MMA or on the endpoints of the interval. So we have the end points of the interval. Those are here zero and two. So what we need to do is we need to find any relative extra MMA that's in that interval as well. So in order to find relative extra MMA we are going to take the derivative of our function. And the excrement occurs either at where the derivative is zero or fails to exist. So we are going to do the derivative of our perimeter function. So we're going to start with P of X equals negative two X squared plus two X plus eight. So the derivative of that would be negative four X plus two. So in order to find the relative extra MMA we are going to set the derivative equal to zero. So we get four X equals two or X equals one half. So now I'm going to set up a chart utilizing my three values of X. My minimum of the interval, my maximum of the interval and the relative extra money I just but shaved. So my perimeter formula again was negative two X squared plus two X plus eight. So I'm going to substitute zero in and I'm gonna get negative two times zero squared plus two times zero plus eight or eight, negative two times a half squared plus two times one half plus eight. And that's going to give me 9.5. And now I'm going to put the two in so I'll get negative two times two squared plus two times two plus eight and that's going to yield a four. So therefore our absolute maximum of Pf X will be 9.5 because this is the highest number and are absolute minimum of P. Of X is for and let's do part C. Now part C is asking you to find the dimensions of the rectangle that yield that largest perimeter. So if you think about what our rectangle looked like, we said this was X. And this was four minus X squared. We just found that X being a half gets us that largest perimeter. So that means mhm. That this is one half and this would be four minus one half squared, or four minus a quarter, leaving us three and three quarters. So our dimensions of the rectangle, yielding that largest perimeter will be one half by three and three quarters. Yeah. And then sure enough, if I do that, if I say one half plus three and three quarters plus one half plus three and three quarters for this side, I will get that area. Now we're ready to start part two of this problem. So we're gonna start with that same image again. So we started with the X. Y axis, we had that part of a parabola again and we had our rectangle fitting inside. And we said this point here was X zero. This point here was X. Y. This point here was 00 at this point here was zero. Why? And I'm going to pull that rectangle out of the picture. So if I were to measure from here to here that my base would be X. And from here up my height would be why? And because I have the function defined as four minus X squared, then I could say that why is the same thing as four minus X squared? So for part A you need to determine a function A. Of X. That represents the area of this rectangle in the area of a rectangle is found by doing length times. With so therefore we will have X times four minus X squared. Or if we distribute that X in we could say four X minus x. Cute! Now we're ready to talk about part B, part B is again applying that extreme value theorem, we need to determine the absolute maximum an absolute minimum and again just to reiterate what the extreme value theorem says is that on a closed interval any continuous function will always have absolute minimum and absolute maximum values. And they will occur either at the end points of that interval. And remember our endpoints of our interval are from 0 to 2 or at any I'm sorry at any relative extra MMA So to find our relative extreme A we are going to have to take our function and take its derivative. So our function is four x minus X cubed. So we're going to find the derivative of that which would end up being four minus three X squared. And we will set that equal to zero. So we could say three x squared equals four X squared equals four thirds X equals plus or minus two over radical three. And when we rationalize that we get two RADICAL 3/3. Now, because of our defined interval from 0 to 2, we are going to end up only using the positive version of this because the negative version doesn't fall in that interval. So again, we're going to utilize a chart. Oops. There we go. And we are going to put our X values the endpoints zero and two along with our relative Ekstrom A of two RADICAL 3/3. Our area formula is for x minus X cubed. So we're going to say four times zero minus zero cubed. And we're gonna get a zero. Well then do four times two radical 3/3 minus two RADICAL 3/3 cubed. And I'm going to just move this to down a little bit that I can show you the work or the solution to this. So we're going to end up with eight radical 3/3 minus eight. Radical 27/27. And Radical 27 will simplify and clean up a little bit. So we can then turn this fraction into 24 Radical 3/27 which again that can clean up a little bit more. If we simplify the 24 27 we will get eight RADICAL 3/9. And when I clean that up, I end up with 16 RADICAL 3/9. And now I'm going to use the other end point to and we will get four times two minus two cubed, which will end up being zero as well. So therefore our absolute maximum of a X. Is 16 Radical 3/9 and our absolute minimum of the function f X would be zero. And finally we're going to do part C and part C. In this problem is asking you two find the dimensions of that rectangle that has the largest area. So again, let's draw our rectangle. We said that this was X and this was for minus X squared and we just found the largest area occurred when X was too radical 3/3. So that means here is too radical 3/3 and here would be for minus two RADICAL 3/3 squared, which would be four minus four times 3/9, which would be for minus four thirds, which would end up being two and two thirds. So the dimensions of the rectangle with the largest area is going to be two radical 3/3. Bye two and two thirds. And that's going to yield the largest area of the 19 radical three over, sorry, the 16 radical 3/9. So, I hope this solution process was helpful.