# Determine the maximum area of a right triangle that has legs equal to x+1 and 2x+4.

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The area of a right triangle with legs x+ 1 and 2x + 4 is (1/2)*(x+1)*(2x +4).

So we have the area given by (1/2)[ 2x^2 + 4x + 2x + 4]

=> (1/2) 2x^2 + 6x + 4

Now it is not possible to find the maximum area of the triangle as the expression x^2 + 3x + 2 grows larger with larger values of x without any limits.

**It is not possible to determine the maximum area of the triangle.**

To determine the maximum area of right triangle that has the legs x+1 and 2x+4 .

The area A(x) of the right angled triangles, whose right angle is formed of the sides of length x+1 and 2x+4 is given by:

A(x) = (1/2)(x+1)(2x+4).

A(x) = (1/2)(2x^2+6x+4).

A(x) is maximum if A'(c) = 0, for which A"(c) < 0.

A'(x) = (1/2))(2x^2+6x+4)'

A'(x) = (1/2){2*2*x+6}.

A'(x) = 0 gives (1/2){4x+6} = 0, Or x = c= -6/4 = -2/3.

A''(x) = (1/2)*4 > 0.

So A(x) is minimum when x = -2/3. So min A(x) = A(-2/3) = (1/2)(-2/3+1)(2*-2/3+4) = (1/2) (1/3)(8/3)= 4/3.

As A(x) = (1/2){2x^2+6x+4}, there is no maximum. A(x) increases unbounded for large x's (both positive and negative).

Since the triangle is a right angle triangle, the area is the half of the product of the legs.

A(x) = (x+1)(2x+ 4)/2

A(x) = 2(x+1)(x+2)/2

We'll simplify and we'll get:

A(x) = (x+1)(x+2)

We'll remove the brackets:

A(x) = x^2 + 3x + 2

The area is maximum for the value of x that represents the root of the first derivative of A(x).

A'(x) = 2x + 3

A'(x) = 0

2x + 3 = 0

2x = -3

x = -3/2

**The area is maximum for the critical value of x = -3/2.**