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10 votes
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A wire of length $340$ mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of $1:2$. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM?

  1. $30$
  2. $40$ 
  3. $120$
  4. $180$
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1 Answer

Best answer
18 votes
18 votes
Let two parts be $x$ and $y$ and part $x$ is used to make square and part $y$ is used to make rectangle.

$\large x + y = 340$

If $x$ part is used to make square, means it is the parameter of square and each side of square is $\frac{x}{4}$

Similarly, for rectangle $2*(l+b) = y$ and $l = 2b$

$2*3b = y \Rightarrow b = \frac{y}{6}$ and $l = \frac{y}{3}$ (we know $y = 340-x$)

Question says the combined area of the square and the rectangle is a MINIMUM

$\color{green}{A = (\frac{x}{4})^2 + \frac{340-x}{6}*\frac{340-x}{3}}$

For combined area to be minimum $\frac{\text{d}A}{\text{d}y} = 0$

$\Rightarrow \frac{2x}{16} - \frac{2(340-x)}{18} = 0$

$\large \Rightarrow x = 160$

But question asks for MINIMUM length of side of square that is $\color{maroon}{\frac{x}{4} = 40mm}$

Correct Answer: $B$
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