# Thomas Simpson and Maxima and Minima - Maximizing area of right triangles

Author(s):
Michel Helfgott

Of all right-angled plane triangles having the same given hypotenuse, to find that whose area is the greatest (Example IV, page 17).

Let $$a$$ be the hypotenuse and $$x, y$$ the legs. We have

$$y = \sqrt{a^2 - x^2}$$, so $$area(x) = \frac{x}{2}\sqrt{a^2 - x^2}$$

Let $$f(x) = \frac{x^2}{4}(a^2 - x^2)$$, the square of the area. Then $$0 = f'(x) = \frac{2a^2}{4}x – x^3$$, consequently $$x = \sqrt{\frac{a}{2}}$$. It follows that

$$y = \sqrt{a^2 - \frac{a^2}{2}} = \frac{a}{\sqrt{2}}$$

Thus, the best we can do is to choose the isosceles right triangle with given hypotenuse a.

Simpson provides an alternative way of solving the problem at hand (he says “the same otherwise”): Assuming that $$y$$ is a function of $$x$$, implicit differentiation—a term not used by the author—leads to $$2x + 2yy' = 0$$ since $$x^2 + y^2 = a^2$$, thus $$y' = -\frac{x}{y}$$ . On the other hand $$area(x) = \frac{1}{2}xy(x)$$, so $$area'(x) = \frac{1}{2}(y + xy')$$. We are looking for the maximum of the area function, thus we have to make $$area'(x) = 0$$. Therefore $$\frac{1}{2}(y + xy') = 0$$, which in turn leads to $$y' = -\frac{y}{x}$$. Consequently $$-\frac{x}{y} = -\frac{y}{x}$$, that is to say $$x = y$$. Again we reach the conclusion that the right isosceles triangle, of given hypotenuse $$a$$, encloses the largest area.

Remarks: It might surprise the reader that, in the first approach, Simpson does not take the derivative of the area but of its square. This is a clever choice since the latter is a simple polynomial. As the author writes right before example IV: “It will be proper to observe here, that the value of a quantity, when a maximum or minimum, may oftentimes be determined with more facility by taking the fluxion of some given part, multiple, or power thereof, than from the fluxion of the quantity itself.” Simpson does not, however, mention that a simple algebraic alternative exists, due to the fact that $$f(x)$$ happens to be a biquadratic function. Indeed $$-\frac{1}{4}x^4 + \frac{a^2}{4}x^2$$ can be transformed into $$-\frac{1}{4}z^2 + \frac{a^2}{4}z$$ by defining $$z = x^2$$. Completing squares, we can tell that the function in the variable $$z$$ adopts its minimum at

$$z = \frac{\frac{-a^2}{4}}{2(\frac{-1}{4})} = \frac{a^2}{2}$$.

So the original equation attains its minimum at

$$x = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$$.

Editor’s Note: This article was published in 2005.