Pergunta
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
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YedaElite · Tutor por 8 anos
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To find the largest rectangle that fits inside the graph of the parabola $y=x^{2}$ below the line $y=a$, we need to determine the dimensions of the rectangle that maximize its area.<br /><br />Let's denote the x-coordinate of the vertex of the parabola as $x_0$. The equation of the parabola is $y=x^{2}$, so the vertex is at $(0,0)$. The line $y=a$ intersects the parabola at two points, $x_1$ and $x_2$, where $x_1$ and $x_2$ are the x-coordinates of the points of intersection.<br /><br />The rectangle that fits inside the parabola and below the line $y=a$ has its top side on the line $y=a$ and its bottom side on the parabola $y=x^{2}$. The width of the rectangle is $x_2 - x_1$, and the height is $a - x_1^2$.<br /><br />The area of the rectangle is given by the product of its width and height, which is:<br /><br />$A = (x_2 - x_1)(a - x_1^2)$<br /><br />To maximize the area, we need to find the values of $x_1$ and $x_2$ that maximize this expression. Taking the derivative of $A$ with respect to $x_1$ and setting it equal to zero, we get:<br /><br />$\frac{dA}{dx_1} = (a - 3x_1^2)(x_2 - x_1) - (x_2 - x_1)(2x_1) = 0$<br /><br />Solving this equation for $x_1$, we find:<br /><br />$x_1 = \frac{a}{3}$<br /><br />Substituting this value back into the equation for $x_2$, we get:<br /><br />$x_2 = \sqrt{a - x_1^2} = \sqrt{\frac{2a}{3}}$<br /><br />Therefore, the dimensions of the largest rectangle that fits inside the graph of the parabola $y=x^{2}$ below the line $y=a$ are:<br /><br />Width: $x_2 - x_1 = \sqrt{\frac{2a}{3}} - \frac{a}{3}$<br />Height: $a - x_1^2 = a - \left(\frac{a}{3}\right)^2 = \frac{2a}{3}$<br /><br />The maximum area of the rectangle is:<br /><br />$A = (x_2 - x_1)(a - x_1^2) = \left(\sqrt{\frac{2a}{3}} - \frac{a}{3}\right)\left(\frac{2a}{3}\right) = \frac{2a^2}{3} - \frac{2a^3}{9}$
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