# 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.

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...

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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.`

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