Step 1

Step 2

Let x and y be the side lengths of the rectangle, then the area is

\(\displaystyle{A}={x}{y}\)

and by similar triangles (ABC and FBE, note: BF is \(\displaystyle{3}-{y}\))

\(\displaystyle{\frac{{{x}}}{{{4}}}}={\frac{{{3}-{y}}}{{{3}}}}\)

\(\displaystyle{\frac{{{3}{x}}}{{{4}}}}={3}-{y}\)

\(\displaystyle{\frac{{{3}{x}}}{{{4}}}}-{3}=-{y}\)

\(\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{x}\)

Step 3

Substitute for y in the area equation, find \(\displaystyle{A}'\)

\(\displaystyle{A}={x}{\left({3}-{\frac{{{3}}}{{{4}}}}{x}\right)}={3}{x}-{\frac{{{3}}}{{{4}}}}{x}^{{{2}}}\)

\(\displaystyle{A}'={3}-{\frac{{{3}}}{{{2}}}}{x}\)

Find where it is 0

\(\displaystyle{0}={3}-{\frac{{{3}}}{{{2}}}}{x}\)

\(\displaystyle{\frac{{{3}}}{{{2}}}}{x}={3}\)

\(\displaystyle{x}={2}\)

Step 4

Find te corresponding y (with the similar triangle equation)

\(\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{\left({2}\right)}={3}-{\frac{{{3}}}{{{2}}}}={\frac{{{6}-{3}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}\)

So the maximum rectangle area is

\(\displaystyle{A}={2}\cdot{\frac{{{3}}}{{{2}}}}={3}{c}{m}^{{{2}}}\)

Step 2

Let x and y be the side lengths of the rectangle, then the area is

\(\displaystyle{A}={x}{y}\)

and by similar triangles (ABC and FBE, note: BF is \(\displaystyle{3}-{y}\))

\(\displaystyle{\frac{{{x}}}{{{4}}}}={\frac{{{3}-{y}}}{{{3}}}}\)

\(\displaystyle{\frac{{{3}{x}}}{{{4}}}}={3}-{y}\)

\(\displaystyle{\frac{{{3}{x}}}{{{4}}}}-{3}=-{y}\)

\(\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{x}\)

Step 3

Substitute for y in the area equation, find \(\displaystyle{A}'\)

\(\displaystyle{A}={x}{\left({3}-{\frac{{{3}}}{{{4}}}}{x}\right)}={3}{x}-{\frac{{{3}}}{{{4}}}}{x}^{{{2}}}\)

\(\displaystyle{A}'={3}-{\frac{{{3}}}{{{2}}}}{x}\)

Find where it is 0

\(\displaystyle{0}={3}-{\frac{{{3}}}{{{2}}}}{x}\)

\(\displaystyle{\frac{{{3}}}{{{2}}}}{x}={3}\)

\(\displaystyle{x}={2}\)

Step 4

Find te corresponding y (with the similar triangle equation)

\(\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{\left({2}\right)}={3}-{\frac{{{3}}}{{{2}}}}={\frac{{{6}-{3}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}\)

So the maximum rectangle area is

\(\displaystyle{A}={2}\cdot{\frac{{{3}}}{{{2}}}}={3}{c}{m}^{{{2}}}\)