# Is the answer 1? I got it wrong? ?

A rectangle is inscribed in a right isosceles triangle with a hypotenuse of length 1 units.

What is the largest area the rectangle can have?

Area=_____units^2

Relevance

if length of hypotenuse of the right isosceles triangle is 1, the the remain sides are ½√2 each.

constructive the right isosceles triangle as bounded by line y = x

if height of rectangle = y then length of rectangle = ( ½√2 - x )

Area of rectangle inscribed, A:

A = ( ½√2 - x )( y )

A = ( ½√2 - x )( x ) since y = x

-A = x² - ½x√2

let’s complete the square

-A = ( x - ¼√2 )² - ( ¼√2 )²

A = ⅛ - ( x - ¼√2 )²

A is maximum if the term ( x - ¼√2 )² is 0;

∴

A = ⅛ - ( x - ¼√2 )²

A = ⅛ - 0

A = ⅛ square units

• Hypotenuse of 1 means the sides are 1/√2

Suppose the rectangle is drawn in the right-angled corner. The opposite corner touches the hypotenuse.

|\

| \

| _\

| |\

|__|_\

x

The small triangles have the same angles as the big triangle, so they similarly have height = width. This makes the height of the rectangle (1/√2-x).

Area = (1/√2 - x)x =  x/√2 - x^2

Set the derivative equal to zero.

1/√2 - 2x = 0

x = 1 / 2√2

Area = 1/8

And the second derivative is -2 everywhere, so this is a maximum.

We could instead draw the rectangle with its side along the hypotenuse.

|\

| \

|/ \ x

|\ . \

|_\/_\

If this rectangle has length x along the hypotenuse, then after some geometry you'll find its area is:

A = (1-x)x/2

Which also has a maximum of 1/8

• If the hypo is 1 then the base and height are each √2/2. Hence the area of the whole triangle is .... 1/4. Convince yourself of this. So the inscribed rectangle must have an area smaller than that! So, try again....