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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

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Konrad509idn
[tex]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(\dfrac{x_2-x_1}{2}\right)\\ x(30-x)=0\\ x=0 \vee x=30\\ A_{max}=A\left(\dfrac{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} [/tex]
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