The dimensions that would result to maximum area will be found as follows: let the length be x, the width will be 32-x thus the area will be given by: P(x)=x(32-x)=32x-x² At maximum area: dP'(x)=0 from the expression: P'(x)=32-2x=0 solving for x 32=2x x=16 inches thus the dimensions that will result in maximum are is length=16 inches and width=16 inches
Therefore, the best and most correct answer among the choices provided by the question is the fourth choice "15 - 2x ≥ 4y". I hope my answer has come to your help.
