Consider the following statement.Of all the rectangles with a given area, the one with smallest perimeter is a square.The following is a proposed proof for the given statement.Let the rectangle have sides x and y and area A, soA = xyory =Ax.The problem is to maximize the perimeter;P(x)=2x +2y =2x +2Ax.NowP(x)=22Ax2=2(x2 A)x2.So the critical number isx = A.SinceP(x)<0for0< x 0forx >A,there is an absolute minimum atx =A.The sides of the rectangle areAandAA=A,so the rectangle is a square.Identify the error(s) in the proposed proof. (Select all that apply.)The first sentence assumes A = xy when this is not the case.The second sentence should say minimize instead of maximize.In the third sentence, the derivative P(x) is incorrect; it should beP(x)=22Ax=2(x A)x.In the fourth sentence, the critical number is incorrect; it should bex =Ainstead of x = A.The fifth sentence should say absolute maximum instead of absolute minimum.The sixth sentence has an incorrect calculation; it should sayAA= Ainstead ofAA=A.(b)Consider the following statement.Of all the rectangles with a given perimeter, the one with greatest area is a square.The following is a proposed proof for the given statement.Let p be the perimeter and x and y the lengths of the sides, sop =2x +2y 2y = p 2x y =12p x.The area isA(x)= x12p x=12px x2.Now settingA(x)=012p 2x =02x =12p x =14p.SinceA(x)=2<0,there is an absolute minimum for A whenx =14pby the second derivative test.The sides of the rectangle are14pand12p 14p =14p,so the rectangle is a square.Identify the error(s) in the proposed proof. (Select all that apply.)In the second sentence, the area A(x) is incorrect; it should beA(x)= x12p x=12px 2x.The third sentence should set A(x)=0 instead of A(x)=0.The third sentence has an incorrect calculation; it should bex =12p.The fourth sentence should say absolute maximum instead of absolute minimum.The fourth sentence should say A(x)=2>0 instead of A(x)=2<0.