Answer:
The answer is "[tex]\bold{\frac{32}{3}}\\[/tex]"
Step-by-step explanation:
The rectangle should also be symmetrical to it because of the symmetry to the y-axis The pole of the y-axis. Its lower two vertices are (-x,0). it means that
and (-x,0), and (x,0). Therefore the base measurement of the rectangle is 2x. The top vertices on the parabola are as follows:
The calculation of the height of the rectangle also is clearly 16-x^2, (-x,16,-x^2) and (x,16,-x^2).
The area of the rectangle:
[tex]A(x)=(2x)(16-x^2)\\\\A(x)=32x-2x^3[/tex]
The local extremes of this function are where the first derivative is 0:
[tex]A'(x)=32-6x^2\\\\32-6x^2=0\\\\x= \pm\sqrt{\frac{32}{6}}\\\\x= \pm\frac{4\sqrt{3}}{3}\\\\[/tex]
Simply ignore the negative root because we need a positive length calculation
It wants a maximum, this we want to see if the second derivative's profit at the end is negative.
[tex]A''\frac{4\sqrt{3}}{3} = -12\frac{4\sqrt{3}}{3}<0\\\\2.\frac{4\sqrt{3}}{3}= \frac{8\sqrt{3}}{3}\\\\\vertical \ dimension\\\\16-(\frac{4\sqrt{3}}{3})^2= \frac{32}{3}[/tex]
