document.write( "Question 40032: Bob has 3000 feet of fencing available to enclose a rectangular field.
\n" );
document.write( "A. express the area A of rectangle as a function of x where x is the length of rectangle.
\n" );
document.write( "B. for what value of x is the area largest?
\n" );
document.write( "C. what is the maximum area? \n" );
document.write( "
Algebra.Com's Answer #25475 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! > 2. Bob has 3000 feet of fencing available to enclose a rectangular field. \n" ); document.write( "> \n" ); document.write( "> a. express the area A of rectangle as a function of x where x is the length \n" ); document.write( "> of rectangle. \n" ); document.write( "L=X...PERIMETER=L+B+L+B=2(L+B)=2(X+B)=3000 \n" ); document.write( "X+B=1500 \n" ); document.write( "B=1500-X \n" ); document.write( "AREA =A= L*B=X(1500-X) \n" ); document.write( "> b. for what value of x is the area largest? \n" ); document.write( "A=X(1500-X)=-(X^2-1500X)=-(X^2-2*X*750+750^2)+750^2 \n" ); document.write( " A=750^2-(X-750)^2 \n" ); document.write( "HENCE THIS WILL BE MAXIMUM WHEN X=750 \n" ); document.write( "> c. what is the maximum area. \n" ); document.write( "> MAXIMUM AREA =750^2-0=750^2=562500 \n" ); document.write( " |