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

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There cannot be a limit placed on how large a triangle can be when only the length of the legs are given and they are in terms of positive values of x.

As the value of x increases, both the legs grow longer and the area of the triangle increases. This can go on forever.

Only the minimum area is constrained by the fact that it cannot be negative.

The right triangle has legs with dimensions x+1 and 2x+4. The area of a right triangle with legs b and h is A = (1/2)*b*h.

For the given triangle the expression for the area is A = (1/2)*(x+1)*(2x+4)

= (x+1)*(x+2)

= x^2 + 3x + 2

Now the first derivative of A is (x^2 + 3x + 2)' = 2x + 3 and the second derivative is 2.

Notice that 2 is a positive number. This shows that A has an extreme point that is a minimum. There is no maximum value for A. This is further proved by the following graph that represents y = x^2 + 3x + 2

Of course the area of a triangle cannot be negative, as a result the value of x cannot lie in the set (-2, -1)

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 has an extreme value for the critical value of x = -3/2.**