Question

A rectangle is constructed with its base on the​ x-axis and two of its vertices on the parabola yequals25minusxsquared. What are the dimensions of the rectangle with the maximum​ area? What is the​ area?

Answer 1

Answer:

The answer is "$\bold{\frac{32}{3}}$"

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:

$A(x)=(2x)(16-x^2)\\\\A(x)=32x-2x^3$

The local extremes of this function are where the first derivative is 0:

$A'(x)=32-6x^2\\\\32-6x^2=0\\\\x= \pm\sqrt{\frac{32}{6}}\\\\x= \pm\frac{4\sqrt{3}}{3}$

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.

$A''\frac{4\sqrt{3}}{3} = -12\frac{4\sqrt{3}}{3}$