Answer
a) If the length of half the base of the beam is one-third of a foot, then the cross-sectional area of the beam is approximately $1.26$ ft$^2$
b) If the length of half the base of the beam is one-half of a foot, then the cross-sectional area of the beam is approximately $1.73$ ft$^2$
c) If the length of half the base of the beam is two-third of a foot, then the cross-sectional area of the beam is approximately $1.99$ ft$^2$
Work Step by Step
Given $A(x)=4x\sqrt{1-x^2}$
a) If the length of half the base of the beam is one third of a foot, then the cross-sectional area of the beam is calculated as below
$A(\frac{1}{3})= 4.\frac{1}{3}\sqrt{1-(\frac{1}{3})^2}
=\frac{4}{3}\sqrt{\frac{8}{9}}=\frac{4}{3}.\frac{2\sqrt2}{3}
=\frac{8\sqrt2}{9}\approx1.26$ ft$^2$
b) If the length of half the base of the beam is one-half of a foot, then the cross-sectional area of the beam is calculated as below
$A(\frac{1}{2})= 4.\frac{1}{2}\sqrt{1-(\frac{1}{2})^2}
=2\sqrt{\frac{3}{4}} =2\frac{\sqrt3}{2}
=\sqrt3 \approx1.73$ ft$^2$
c) If the length of half the base of the beam is two-third of a foot, then the cross-sectional area of the beam is calculated as below
$A(\frac{2}{3})= 4.\frac{2}{3}\sqrt{1-(\frac{2}{3})^2}
=\frac{8}{3}\sqrt{\frac{5}{9}}=\frac{8}{3}.\frac{\sqrt5}{3}
=\frac{8\sqrt5}{9}\approx1.99$ ft$^2$
