Rectangle Area Change 632E72

1. **Problem statement:** Find the percentage change in the area of a rectangle whose one side is four times the other, given that the perimeter decreases by 2%. 2. **Define variables:** Let the shorter side be $x$. Then the longer side is $4x$. 3. **Formulas:** Perimeter $P = 2(x + 4x) = 2(5x) = 10x$. Area $A = x \times 4x = 4x^2$. 4. **Relate changes:** Given the perimeter decreases by 2%, so $$\frac{dP}{P} = -0.02$$ 5. **Find $dx$ in terms of $dP$:** Since $P = 10x$, differentiate: $$dP = 10 \, dx$$ Divide both sides by $P=10x$: $$\frac{dP}{P} = \frac{10 \, dx}{10x} = \frac{dx}{x}$$ So, $$\frac{dx}{x} = \frac{dP}{P} = -0.02$$ 6. **Find percentage change in area:** Differentiate area: $$dA = 8x \, dx$$ Divide both sides by $A=4x^2$: $$\frac{dA}{A} = \frac{8x \, dx}{4x^2} = 2 \frac{dx}{x}$$ Substitute $\frac{dx}{x} = -0.02$: $$\frac{dA}{A} = 2 \times (-0.02) = -0.04$$ 7. **Interpretation:** The area decreases by 4%. **Final answer:** The area decreases by 4% when the perimeter decreases by 2%.