## Algebra and Trigonometry 10th Edition

$x=25~ft$ $y=\frac{100}{3}~ft$ $A=5000~ft^2$
We need to find the vertex of $A=-\frac{8}{3}x^2+\frac{400}{3}x$ $A=-\frac{8}{3}x^2+\frac{400}{3}x~~$ ($a=-\frac{8}{3},b=\frac{400}{3}x,c=0$): $-\frac{b}{2a}=-\frac{\frac{400}{3}}{2(-\frac{8}{3})}=-\frac{400}{3}(-\frac{3}{16})=25$ $f(25)=-\frac{8}{3}(25)^2+\frac{400}{3}(25)=\frac{5000}{3}$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(25,\frac{5000}{3})$ That is, when $x=25~ft$, $A=\frac{5000}{3}~ft^2$ $y=\frac{200-4x}{3}~~$ (item(a)) $y=\frac{200-4(25)}{3}=\frac{100}{3}~ft$