3. Find the largest rectangle (that is , the rectangle with largest area) that fits inside the graph of the parabola y=x^2 below the line y=a (a is an unspecified constant value), with the top side of the rectangle on the horizontal line y=a

To find the largest rectangle that fits inside the graph of the parabola $y=x^{2}$ below the line $y=a$, we need to maximize the area of the rectangle.<br /><br />Let's denote the x-coordinate of the vertex of the rectangle as $x_0$. Since the rectangle is symmetric about the y-axis, its top side will be on the horizontal line $y=a$ and its bottom side will be on the parabola $y=x^{2}$. Therefore, the height of the rectangle is $a-x_0^2$.<br /><br />The width of the rectangle is twice the distance from the y-axis to the x-coordinate of the vertex, which is $2x_0$.<br /><br />The area of the rectangle is given by the product of its width and height, which is $A = 2x_0(a-x_0^2)$.<br /><br />To maximize the area, we need to find the value of $x_0$ that maximizes the function $A(x_0) = 2x_0(a-x_0^2)$. We can do this by taking the derivative of $A(x_0)$ with respect to $x_0$ and setting it equal to zero:<br /><br />$A'(x_0) = 2(a - 3x_0^2) = 0$<br /><br />Solving for $x_0$, we get $x_0 = \sqrt{a/3}$.<br /><br />Therefore, the largest rectangle that fits inside the graph of the parabola $y=x^{2}$ below the line $y=a$ has a width of $2\sqrt{a/3}$ and a height of $a - (\sqrt{a/3})^2 = 2a/3$.