Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

Explanation

Let one side be `x`, then the other side is `sqrt{25−x^2}.`

Area of triangle `=A=1/2xsqrt{25-x^2}`

`{dA}/{dx}=1/2{25-2x^2}/{sqrt{25-x^2}}`

Put `{dA}/{dx}=0`

Thus `1/2{25-2x^2}/{sqrt{25-x^2}}=0 `

`implies 25-2x^2=0`

`implies x=5/sqrt2`(as side cannot be negative)

And `{d^2A}/{dx^2}={2x^3-75x}/{2(25-x^2)sqrt{25-x^2}}`

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