Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral.
$s=b$ so the triangle is equilateral
were asked to show that up. All the isosceles triangles with a given perimeter the one with the greatest area is equal lateral. Really? So to understand this problem, we know that isosceles triangle can be drawn sort of like this. So we have two sides s with the same length and then a base with a different length or maybe the same length and then a height h from the base to the top. Now we can express the height h in terms of s and be with the Pythagorean theorem. Yeah, yes. So we have The height s is equal to the square of the hypotenuse s squared minus B over two, A squared. Okay, Now the area of the whole triangle is half the base times the height. Yes. And then we want to write area in terms of b. Sorry. Before we even do this, this is one half B times the square root of X squared minus. This is B squared over four. Great. Yeah, and we'll let p be the given perimeter Soapy is actually a constant Well, the perimeter is given by two s plus B and so we can write he as a function or s as a function of P and B s is equal to P minus B over to and so plugging this in we get the area as a function of be strictly is one half B times the square root of P minus B squared over four minus B squared over four and this reduces eventually to 1/4 B times the square root of P squared minus two PB to buy super too. This one now, in order to maximize the area, will find the derivative a prime of B. It is true. Thank you. Mhm. Now, on the one hand, you could take the derivative of this expression. Alternatively, we know that the area is also maximized right when the area squared, which I'll call f is maximized to so f of b is B squared over 16 times p squared minus two p. B, which is P squared over 16 times B squared minus two p over 16, which is p over eight. Pretty sick. Just be cubed. Yeah, the name's shoes. Yeah. Yes. So the derivative of crime of B, this is P squared over eight B minus three p over eight b squared we want to find when this is equal to zero, we can factor this as yeah. Factor at API. Over eight b times P minus three b equals zero I love and so we get that be equal Zero or B equals p over three. Now. Very clearly B equals zero is not a reasonable answer. It's not going to be, yes, maximum area. So we only have the critical value of B equals P over three. No. Now, looking back at our area function, we see that mhm or even just looking at F prime of B F prime of B is going to be greater than zero for be less than p over three in the F prime of B is less than zero for be greater than P over three. You can see this from our factory ization of F B. Therefore, it follows that F has a maximum value at B equals P over three. By our previous discussion, it follows that a also has a maximum value at B equals P over three. Substituting into our perimeter equation, we have two times s plus p over three equals p. So s is equal to p over three, but then it follows. That s is equal to be since s is equal to be. All three sides are the same. It follows that our triangle is an equal lateral triangle. Yeah, these signatures officially used part was the second.