Question

A right triangle has one vertex on the graph of y = -(1/2)x^2 + 10, x > 0, at (x, y), another at the origin, and the third on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x. For what value of x is A the largest?

Answer 1

The area A of the right triangle as a function of x is A(x) = (1/2) * x * (-(1/2)x^2 + 10). To find the x-value that maximizes A, differentiate A with respect to x and set the derivative equal to zero.

Step-by-step explanation:

The area A of the right triangle can be expressed as a function of x using the base and the height of the triangle. With one vertex on the curve y = -(1/2)x^2 + 10, we know the height of the triangle (y-coordinate) at a given x is y = -(1/2)x^2 + 10. The base of the triangle is along the x-axis, from the origin to (x, 0), hence the base has length x.

The area of a triangle is given by Area = 1/2 * base * height. Hence, the area A as a function of x is:

A(x) = (1/2) * x * (-(1/2)x^2 + 10)

To find the value of x that maximizes the area A, we can differentiate A with respect to x and find the critical points. Setting the derivative equal to zero will give us the value(s) of x where the area is at a maximum. Solving A'(x) = 0 will give us the desired x-value.

The differentiation gives A'(x) = (1/2)(-3/2)x^2 + 10. Setting this equal to zero and solving for x yields the maximum area at certain x-value.