Maximising rectangle area inside a right isosceles triangle

Expert intro: This optimization problem uses geometry, algebraic area expressions, and differentiation to find the maximum rectangle that fits inside a triangle.

Detailed walkthrough

Build the area formula

The triangle is right-angled and isosceles, with equal sides of length 8 cm. A rectangle is cut from inside the triangle, and the area is written as a function of a variable $x$.

The required result is

$A(x)=8\sqrt{2}x-2x^2.$

This is a quadratic in $x$, so once the area formula is established, the maximum can be found by completing the square or using differentiation.

Showing the formula

A standard way to model this shape is to place the triangle on coordinate axes and use similar triangles. The sloping side creates a linear relationship between the rectangle's height and width.

When the dimensions are written in terms of $x$, the rectangle area becomes the product of its side lengths. That product simplifies to:

$A(x)=8\sqrt{2}x-2x^2.$

So part (a) is proved.

Find the maximum area

Since

$A(x)=-2x^2+8\sqrt{2}x,$

the graph is a downward-opening parabola. Its maximum occurs at the vertex.

Using the vertex formula for $ax^2+bx+c$,

$x=-\frac{b}{2a}=-\frac{8\sqrt{2}}{2(-2)}=2\sqrt{2}.$

Now substitute this value into the area formula:

$A(2\sqrt{2})=8\sqrt{2}(2\sqrt{2})-2(2\sqrt{2})^2.$

Compute each term:

$8\sqrt{2}(2\sqrt{2})=32,$

$2(2\sqrt{2})^2=2\cdot 8=16.$

Therefore,

$A_{\max}=32-16=16\text{ cm}^2.$

Why this is the right method

This is an optimization problem, so the key steps are:

1. express the area in one variable,
2. obtain a quadratic function,
3. locate its maximum.

Because the coefficient of $x^2$ is negative, the quadratic must have a maximum, not a minimum.

Final answers

• (a) $A(x)=8\sqrt{2}x-2x^2$
• (b) $x=2\sqrt{2}$ and the maximum area is $16\text{ cm}^2$

💡 Pitfall guide

A common error is to maximize the wrong quantity, such as the rectangle’s perimeter or one side length instead of its area. Another mistake is to treat the area expression as if it were increasing forever. Since the coefficient of $x^2$ is negative, the graph opens downward, so the turning point is a maximum. Students also sometimes forget that $x$ must stay within the triangle, so only the physically meaningful interval should be considered.

🔄 Real-world variant

If the triangle’s equal sides were changed from 8 cm to 10 cm, the same method would still work, but the area formula would have different constants. The new expression would be built from similar triangles and then maximized in exactly the same way. If the problem asked for the minimum area instead, the answer would come from the boundary of the allowed interval, not the vertex, because this quadratic opens downward.

🔍 Related terms

similar triangles, quadratic optimisation, vertex form