# Find the area of the largest rectangle that can be inscribe in a right triangle with legs of length 3cm and 4cm, if the two sides of the rectangle lie along the legs.

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### 1 Answer

You should use the following notation for the length and width of rectangle such that: l-length and w-width.

Notice that you may evaluate the area of right angle triangle such that:

`A = (3*4)/2 => A = 6 cm^2`

You also may evaluate the area of triangle adding the areas of two small triangles and rectangle within such that:

`A = (4-w)*l/2 + w*l + (3-l)*w/2`

`A = 2l - (wl)/2 + wl + 3w/2 - (wl)/2`

`A = 2l + 3w/2`

Substituting `6 cm^2` for the area of triangle yields:

`6 = 2l + 3w/2 => 12 = 4l + 3w => w = (12 - 4l)/3`

You need to evaluate the area of rectangle such that:

`A_1 = w*l`

You need to write the area of rectangle in terms of length only such that:

`A_1 = ((12 - 4l)/3)*l`

`A_1 = 4l - 4l^2/3`

You need to optimize this area differentiating the expression with respect to l such that:

`(A_1)' = 4 - 8l/3`

You need to solve for l the equation `4 - 8l/3 = 0` such that:

`4 - 8l/3 = 0 => 12 - 8l = 0 => 12 = 8l => l = 12/8 => l = 3/2 cm`

`w = (12 - 4l)/3 =>w = (12 - 6)/3 => w = 2 cm`

**Hence, evaluating the area of the largest rectangle that can be inscribed in the given right angle triangle yields `A_1 = 2*(3/2) = 3 cm^2.` **