Applications of Derivatives
Maxima/Minima
MMTS_Full_Test_01
Grade 12
Question:
A wire of length 36cm is cut into two pieces. One piece is bent into a square and the other into an equilateral triangle. The minimum value of the total area is
$\dfrac{324}{4+9\sqrt{3}}$ sq.cm
$\dfrac{164}{4+\sqrt{3}}$ sq.cm
$\dfrac{108\sqrt{3}}{4+9\sqrt{3}}$ sq.cm
Both 1 and 3
Step-by-Step Solution
Key Concept: Let $x$ be length for square; $36-x$ for equilateral triangle; minimize total area
$A=\frac{x^2}{16}+\frac{(36-x)^2\sqrt{3}}{36}$. $A'=\frac{x}{8}-\frac{(36-x)\sqrt{3}}{18}=0\Rightarrow x=\frac{144\sqrt{3}}{18+8\sqrt{3}}=\frac{144\sqrt{3}}{9+4\sqrt{3}}\cdot\frac{1}{2}$. Min area $=\frac{324\sqrt{3}}{4+9\sqrt{3}}$... answer: 3.
Correct Answer: 3