### Problem Statement:Suppose that 404 feet of fencing are used to enclose a corral in the shape of a rectangle with a semicircle, whose diameter is a side of the rectangle, as shown in the following figure:**Diagram Explanation:**The diagram illustrates a rectangular corral with a semicircle on top. The diameter of the semicircle is the top side of the rectangle. The width of the rectangle is labeled as "x" and the height as "y".### Objective:Find the dimensions of the corral that will yield the maximum area.**Inputs for Solution:**- \( x = \) ____ ft.- \( y = \) ____ ft.

Area of the shape=Area of the semicircle+Area of rectangle .Area of semicircle=πx22x12 =πx28Area of rectangle=xySo , Area of the shap=πx28+xy Fencing=404 ftmeans , Perimeter=404⇒πx2+x+2y=404So , we want to maximize πx28+xysubject to π2+1x+2y=404⇒2y=404-π2+1x⇒y=202-π4+12xSo πx28+xy=πx28+x202-π4+12x =πx28+202x-π4+12x2 =gx sayNow g'x=2xπ8+202-2π4+12x =xπ4+202-π2+1xPut g'x=0⇒xπ4-π2+1x+202=0 So , π4-π2-1x=-202⇒π-2π-44x=-202⇒-π-44x=-202⇒x=808π+4=8087.14=113.17 feety=202-π4+12113.16 =202-π+24x113.16=202-581.644=202-145.41 =54.58 feet