Applications of Derivatives
Constrained Extrema
Grade 12
Question:
<p>A wire of length 2 units is cut into two parts which are bent respectively to form a square of side <i>x</i> units and a circle of radius <i>r</i> units. If the sum of the areas of the square and the circle so formed is minimum, then:</p>
<p>(a) \(2x = (\pi + 4)r\)</p>
<p>(b) \((4 - \pi)x = \pi r\)</p>
<p>(c) \(x = 2r\)</p>
<p>(d) \(2x = r\)</p>
Step-by-Step Solution
Key Concept: Use the perimeter constraint to relate x and r, then minimize the total area by calculus.
<p>The perimeter constraint is: $4x + 2\pi r = 2$. The total area is $A = x^2 + \pi r^2$. Using the constraint to express $x$ in terms of $r$ and minimizing $A$ with respect to $r$, we differentiate and set $\frac{dA}{dr} = 0$. This yields the relationship $2x = (\pi + 4)r$.</p>
Correct Answer: a