# 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. - Mathematics

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

Concept: Simple Problems on Applications of Derivatives
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