The sum of the lengths of the legs of a right triangle is 40cm. What is the largest possible area for such triangle?
Area = ·base·height
Let the area by y, one leg be x and the other leg be 40-x, then
y = x(40-x)
y = 40x - x²
y = 20x - x²
Write that in the form
y = ax² + bx + c
y = x² + 20x + 0
Since the coefficient of x² is negative, we know that the
parabola opens downward and therefore the vertex will be a
maximum.
The vertex of a parabola is given by
x-coordinate of the vertex =
y-coordinate of the vertex = what you get when you substitute
the x-coordinate in the equation
and solve for y.
So for your problem:
x-coordinate of the vertex = = = 20
y-coordinate of the vertex = y = (20)² + 20(20) + 0 = 200
So the maximum area that such a right triangle can have is 200
and that will be when x = 20, which means that the other leg 40-x
will also be 40-20 = 20. This triangle will be an isosceles right
triangle.
Answer: Maximum area = 200cm² when both legs are 20cm each.
Edwin