Among all isosceles triangles of perimeter 10, which one has the largest area?

The equilateral triangle with all sides having a length of 10/3 has the largest possible area of all isosceles triangles of perimeter 10.
    The area of a triangle is 1/2 bh where b is the base and h is the height of the triangle. So let's create an equation for the area of an isosceles triangle with a perimeter of 10, given the base.
   So with a specified base, the length of the other 2 sides will be: (10 - b)/2
And using the Pythagorean theorem, the height of the triangle will be: sqrt(((10 - b)/2)^2 - (b/2)^2)
   So the area of the triangle will be: A = b*sqrt(((10 - b)/2)^2 - (b/2)^2)/2

Let's simplify the equation a bit:
 A = b*sqrt(((10 - b)/2)^2 - (b/2)^2)/2 A = b*sqrt( (10 - b)^2/4 - b^2/4 )/2
 A = b*sqrt( (100 - 20b + b^2)/4 - b^2/4 )/2
 A = b*sqrt(25 - 5b + b^2/4 - b^2/4 )/2
 A = b*sqrt(25 - 5b)/2
 A = (1/2)b*sqrt(25 - 5b)

   Now whenever you see "largest", you should think of "first derivative". So let's calculate that. A = (1/2)b*sqrt(25 - 5b) A = (1/2)b*(25 - 5b)^(1/2)

   Using the product rule we get: A' = (1/2)*(25 - 5b)^(1/2) + (1/2)b*(1/2)(25 - 5b)^(-1/2)(-5)
   Simplify:
 A' = (1/2)*(25 - 5b)^(1/2) - (5/4)b*(25 - 5b)^(-1/2)
 A' = (25 - 5b)^(1/2))/2 - (5b)/(4(25 - 5b)^(1/2))
 A' = (25 - 5b)^(1/2))/2 * (2(25 - 5b)^(1/2))/(2(25 - 5b)^(1/2)) - (5b)/(4(25 - 5b)^(1/2))
 A' = 2(25 - 5b)//(4(25 - 5b)^(1/2)) - (5b)/(4(25 - 5b)^(1/2))
 A' = (2(25 - 5b) - 5b)/(4(25 - 5b)^(1/2))
 A' = (50 - 10b - 5b)/(4(25 - 5b)^(1/2))
 A' = (50 - 15b)/(4(25 - 5b)^(1/2))
   Now the only possible values of b that will be a local minimum or maximum will be those points where the first derivative is either 0, or undefined. It will have the value 0 when the term (50-15b) is 0, or it will be undefined when the bottom term 4(25-5b)^(1/2) is 0. So let's check those possible values: 50 - 15b = 0 50 = 15b 50/15 = b 10/3 = b
   So if the base is 10/3, that means that the other 2 sides of the isosceles triangle are (10 - 10/3)/2 = (30/3 - 10/3)/2 = (20/3)/2 = 10/3, which is an equilateral triangle. Let's check the second possibility: 4(25-5b)^(1/2) becomes undefined when b = 5, giving a value of 0. And if the base has a length of 5, that leaves 5 for the combined length of the other 2 sides, when means that you have a degenerated triangle with a height of 0. So the area of that triangle is also 0. Therefore the largest possible isosceles triangle with a perimeter of 10 is an equilateral triangle with all three sides having length 10/3.

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