The smallest perimeter

Good question! Since the minimum perimeter for a set area is when $x = \frac{b}{2}$ , that is true for any base $b$ , and by extension any height that makes $A = \frac{1}{2}bh$ a true statement.

Once you know that the minimum perimeter of a triangle with a given area and a given base occurs when the triangle is an isosceles triangle, another way to finish the solution is to show that the perimeter $P$ of an isosceles triangle with a base $b$ and a height $h$ and a fixed area $A$ is $P = b + 2\sqrt{(\frac{b}{2})^2 + h^2}$ , and since $A = \frac{1}{2}bh$ , $P = b + 2\sqrt{(\frac{b}{2})^2 + (\frac{2A}{b})^2}$ and $P' = 1 + \frac{b^4 - 16A^2}{b^2 \sqrt{b^4 + 16A^2}}$ . Setting $P' = 0$ and solving for $A$ gives $A = \frac{\sqrt{3}}{4}b^2$ , which is the area formula for an equilateral triangle.

If the area and the base are constant, then the perimeter of the equilateral triangle < the perimeter of an (other) acute triangle < the perimeter of the right triangle < the perimeter of the obtuse triangle. However, this may or may not be the case if the base is not constant. For example, the perimeter of a right triangle with sides ( $10\sqrt{2}$ , $10\sqrt{2}$ , $20$ ) is smaller than the perimeter of an acute triangle with sides ( $5\sqrt{17}$ , $5\sqrt{17}$ , $10$ ), even though both have the same area.

The last sentence of my solution deals with that.

When a derivative is equal to zero, then the rate of change is equal to zero, and which are where possible maximum and minimum values occur. In the context of this problem, setting $P' = 0$ and solving for $x$ gives the $x$ value (the length of the side) at which $P$ (the perimeter) is a minimum.

Here are more details on one way to solve $P' = 0$ :

$0 = \frac{x}{\sqrt{x^2 + h^2}} - \frac{b - x}{\sqrt{(b - x)^2 + h^2}}$

$\frac{b - x}{\sqrt{(b - x)^2 + h^2}} = \frac{x}{\sqrt{x^2 + h^2}}$

$(b - x)(\sqrt{x^2 + h^2}) = x(\sqrt{(b - x)^2 + h^2})$

$(b - x)^2(x^2 + h^2) = x^2((b - x)^2 + h^2)$

$(b - x)^2x^2 + (b - x)^2h^2 = x^2(b - x)^2 + x^2h^2$

$(b - x)^2h^2 = x^2h^2$

$(b - x)^2 = x^2$

$b - x = x$

$b = 2x$

$\frac{b}{2} = x$

$x = \frac{b}{2}$