Largest rectangle in triangle

Consider the scalene triangle as shown in the above diagram. The area of this triangle is:

$M = \frac{ab \sin{\theta}}{2}$

The area of the rectangle EFGH is:

$A = (x_1 + x_2)y$

By analysing the above diagram, one can apply basic trigonometric ratios and properties of similar triangles to arrive at the following two relations. The first relation can be found by looking at triangle AFE and the second one is found by exploiting the fact that the triangle GHC is similar to that containing the side BG.

$\frac{y}{a\cos{\theta} - x_1} = \tan{\theta}$ $\frac{a\sin{\theta}-y}{x_2} = \frac{y}{b -x_2-a\cos{\theta}}$

Rearranging the above relations:

$a\cos{\theta} - y\cot{\theta} = x_1$ $\frac{(b-a\cos{\theta})(a\sin{\theta}-y)}{a\sin{\theta}}=x_2$

Plugging these into the formula for the area of the rectangle gives $A$ as a quadratic function of $y$ . Now, since $y$ is the only variable in this function, the maximum value can be found by:

$\frac{dA}{dy}=0$

Solving for the value of $y$ gives the following. Steps of evaluation are left out.

$y_c = \frac{a\sin{\theta}}{2}$

The above value of $y$ corresponds to a maxima which can be checked by a simple second derivative test. The corresponding maximum rectangle area is found by substituting this value of $y$ in $A(y)$ . The result after simplifications is:

$A_{\mathrm{max}} = \frac{ab \sin{\theta}}{4}=\frac{M}{2}$

One can also avoid calculus to find the maximum. The function $A(y)$ is a quadratic function, and its maximum value can be found by completing the squares. Here, one can safely say that the extreme value is maximum since the coefficient of $y^2$ in $A(y)$ is negative.

Try this paper .

Taking the first quadrant of the $xy-$ plane, let the right triangle $AOB$ be defined by the following points:

$A(0,\frac{2M}{k}); O(0,0): B(k,0)$ , where $k \in \mathbb{R^{+}}$

The hypothenuse $AB$ is defined by the line $y = -\frac{2M}{k^2}x + \frac{2M}{k}$ . Now, pick an arbitrary point $P(x_{0}, -\frac{2M}{k^2}x_{0} + \frac{2M}{k})$ (where $0 < x_{0} < k$ ) such that the area of the inscribed rectangle whose sides are parallel with the coordinate axes is the function:

$A(x_{0}) = x_{0}(-\frac{2M}{k^2}x_{0} + \frac{2M}{k}) = -\frac{2M}{k^2}x^{2}_{0} + \frac{2M}{k}x_{0}$ (i).

Setting the first derivative equal to zero yields the critical point:

$A'(x_{0}) = -\frac{4M}{k^2}x_{0} + \frac{2M}{k} = 0 \Rightarrow x_{0} = \frac{k}{2}$ (ii),

and the second derivative at this critical point is:

$A''(k/2) = -\frac{4M}{k^2} < 0$ (iii)

hence a maximum. Thus, the maximum rectangular area is just $A(k/2) = \boxed{\frac{M}{2}}.$