If we draw a rectangle under the graph, we can find that the width w of the rectangle is 2x and the height h is 25 - x^2.
So, the area of the rectangle is,
[tex]\begin{gathered} A=2x(25-x^2) \\ A=50x-2x^3 \\ \end{gathered}[/tex]Differentiate the function to find maxiumum.
[tex]A^{\prime}=50-6x^2[/tex]Putting A'=0,
[tex]\begin{gathered} x^2=\frac{50}{6}=\frac{25}{3} \\ x=+\frac{5}{\sqrt[]{3}}or-\frac{5}{\sqrt[]{3}} \end{gathered}[/tex]x should be positive So,
[tex]x=\frac{5}{\sqrt[]{3}}[/tex]Differentiate A' with respect to x to find if it is a maximum.
[tex]\begin{gathered} A^{\doubleprime}=-12x<0 \\ \text{when x=}\frac{\text{5}}{\sqrt[]{3}} \end{gathered}[/tex]It confirms that area is a maximum for x.
So, the width of the rectangle with largest area is ,
[tex]w=2x=2\times\frac{5}{\sqrt[]{3}}=\frac{10}{\sqrt[]{3}}[/tex][tex]\begin{gathered} h=25-x^2 \\ h=25-(\frac{5}{\sqrt[]{3}})^2 \\ h=25-\frac{25}{3} \\ h=\frac{50}{3} \end{gathered}[/tex]
