Question

Ex 2.10

3) the perimeter of a rectangle is 60m and its length is x m. show that the area of the rectangle is given by the equation A=30x-x². hence find the maximum area of the rectangle

Answer 1

`P=60 \text{ m}=2l+2w\\ l=x \text{ m}\\ A=lw\\\\ 2l+2w=60\\ 2w=60-2l\\ w=30-l \\w=30-x\\\\ A=x\cdot(30-x)\\ A=30x-x^2\\\\ A_(max)=A\left((x_2-x_1)/(2)\right)\\ x(30-x)=0\\ x=0 \vee x=30\\ A_(max)=A\left((30-0)/(2)\right)\\ A_(max)=A(15)\\ A_(max)=30\cdot15-15^2\\ A_(max)=450-225\\ \boxed{A_(max)=225 \text{ m}^2}`