A side of a rectangle is three times the other. If the perimeter increases by 2%, find the percentage increase in the area of the rectangle.
Question 1 Report
A side of a rectangle is three times the other. If the perimeter increases by 2%, find the percentage increase in the area of the rectangle.
Answer Details
Rectangle with one side three times the other; perimeter rises by 2%.
Let the shorter side be \(x\), so the longer side is \(3x\).
\[P=2(x+3x)=8x,\qquad A=x(3x)=3x^{2}\]The perimeter \(P=8x\) is directly proportional to \(x\), so a \(2\%\) rise in perimeter is a \(2\%\) rise in \(x\). Using differentials for the area:
\[A=3x^{2}\;\Rightarrow\;\frac{dA}{A}=2\cdot\frac{dx}{x}\]\[\frac{\Delta A}{A}\times 100\%=2\times 2\%=4\%\]Alternatively, the new side is \(1.02x\), so the new area is \(3(1.02x)^{2}=1.0404\times 3x^{2}\), an increase of \(4.04\%\), which rounds to about \(4\%\).
The area increases by approximately \(4\%\).
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