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Question
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60º.
Solution
Let ABC be the right angled triangle with
base b and hypotenuse h.
Given that b+h=k
Let A be the area of the right triangle.
`A=1/2 xxbxxsqrt(h^2-b^2)`
`A^2=1/4b^2(h^2-b^2)`
`A^2=b^2/4((k-b)^2-b^2) [because h=k-b]`
`A^2=b^2/4(k^2+b^2-2kb-b^2)`
`A^2=b^2/4(k^2-2kb)`
`A^2=(b^2k^2-2kb^3)/4`
``Differentiating the above function with respect to be, we have
`2A (dA)/(db)=(2bk^2-6kb^2)/4.......(1)`
`=>(dA)/(db)=(bk^2-3kb^2)/(2A)`
For the area to be maximum, we have
`(dA)/(db)=0`
`=>bk^2-3kb^2=0`
`bk=3b^2`
`b=k/3`
Again differentiating the function in equation (1), with respect to b, we have
`2((dA)/(db))2+2A(d^2A)/(db^2)=(2k^2-12kb)/4.....(2)`
Now substituting 0 and b in equation (2), we have
`2A(d^2A)/(db^2)=(2k^2-12k(k/3))/4`
`2A(d^2A)/(db^2)=(6k^2-12k^2)/12`
`2A(d^2A)/(db^2)=-k^2/2`
`2A(d^2A)/(db^2)=-k^2/(4A)<0`
Thus area is maximum at b=k/3.
Now, ` h=k-k/3=(2k)/3`
Let be he angle between the base of triangle and hypotenuse of the right triangle.
Thus, `costheta=b/h=(k/3)/((2k)/3)=1/2`
`=>theta=cos^(-1)(1/2)=pi/3`
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