Wait, why do we use $4x + 2\pi r = 2$ instead of just $x + r = 2$?

That is a classic trap! Remember, $x$ is the side of the square and $r$ is the radius of the circle. The wire forms the *boundary* of these shapes. The boundary of a square is its perimeter ($4x$), and the boundary of a circle is its circumference ($2\pi r$). You cannot add side lengths and radii directly because they represent different geometric properties. Always relate the wire length to the perimeter/circumference, not the dimensions themselves.

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Animated Solution for Mathematics - Differentiation: A wire of length $2$ units is cut into two parts which are bent respectively to form a square of side $= x$ units and a circle of radius $= r$ units. If the sum of the areas of the square and the circle so formed is minimum, then

Q: Is it always true that the side of the square equals the diameter of the circle for minimum area?

In this specific problem, yes! When you solve for $x$ and $r$, you will find $x = \frac{2}{\pi + 4}$ and $r = \frac{1}{\pi + 4}$. Since the diameter of the circle is $2r$, we see that $x = 2r$. This is a beautiful, elegant result that often appears in optimization problems involving geometric shapes. It suggests a certain 'balance' between the shapes that minimizes the total area.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the wire as a fixed resource. The total length of the wire is

2

units. When you cut it into two pieces to form a square of side

x

and a circle of radius

r

, the sum of the square's perimeter (

4 x

) and the circle's circumference (

2 π r

) must equal the total length of the wire,

2

. This is your fundamental constraint.

Hint 2

Hint 2 (The Arsenal): You need to express the total area

A = x^{2} + π r^{2}

as a function of a single variable. Use the constraint equation

4 x + 2 π r = 2

to isolate

r

in terms of

x

. This will allow you to substitute

r

into the area formula, turning it into a quadratic function of

x

Hint 3

Hint 3 (The Execution): Once you have

A (x) = x^{2} + π (\frac{1 - 2 x}{π})^{2}

, differentiate

A

with respect to

x

and set the derivative

\frac{d A}{d x}

to zero to find the critical point. Solve this linear equation for

x

, and then find the corresponding

r

to see the relationship between them.

Visualized Solution

Visualizing the Wire Cut

Total wire length = $2$ units
Part 1 forms a Square of side $x$
Part 2 forms a Circle of radius $r$

Establishing the Perimeter Constraint

Perimeter of Square + Circumference of Circle = Total Length
$4 x + 2 π r = 2$

Simplifying the Constraint

Divide the equation by $2$ :
$2 x + π r = 1$

Expressing $r$ in terms of $x$

Isolate $r$ :
$π r = 1 - 2 x$
$r = \frac{1 - 2 x}{π}$

The Total Area Function $A (x)$

Total Area $A = Area_{s q u a r e} + Area_{c i r c l e}$
$A = x^{2} + π r^{2}$

Area in terms of a Single Variable

Substitute $r = \frac{1 - 2 x}{π}$ into $A$ :
$A = x^{2} + π (\frac{1 - 2 x}{π})^{2}$
$A = x^{2} + \frac{( 1 - 2 x ) ^{2}}{π}$

Condition for Minimum Area

To minimize the area, the first derivative must be zero:
$\frac{d A}{d x} = 0$

Differentiating the Area Function

Differentiate $A$ with respect to $x$ :
$\frac{d A}{d x} = 2 x + \frac{1}{π} \cdot 2 (1 - 2 x) \cdot (- 2)$

Setting Derivative to Zero

Set $\frac{d A}{d x} = 0$ :
$2 x - \frac{4 ( 1 - 2 x )}{π} = 0$

Solving for $x$

Multiply by $π$ :
$2 π x - 4 (1 - 2 x) = 0$
$2 π x - 4 + 8 x = 0$
$x (2 π + 8) = 4 \Rightarrow x = \frac{2}{π + 4}$

Finding the Radius $r$

Substitute $x = \frac{2}{π + 4}$ into $r = \frac{1 - 2 x}{π}$ :
$π r = 1 - 2 (\frac{2}{π + 4})$
$π r = \frac{π + 4 - 4}{π + 4}$

Simplifying $r$

$π r = \frac{π}{π + 4}$
$r = \frac{1}{π + 4}$

The Final Relation: $x = 2 r$

Compare $x$ and $r$ :
$x = \frac{2}{π + 4}$ and $r = \frac{1}{π + 4}$
Clearly, $x = 2 r$

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Imagine you are holding a wire exactly

2

units long. We are going to cut this wire into two pieces to form a square of side

x

and a circle of radius

r

This is a study of how we distribute a finite resource to minimize the total area occupied by these two shapes.

The Constraint Equation

The total length of the wire is fixed at

2

units. The perimeter of the square is

4 x

, and the circumference of the circle is

2 π r

The sum of these lengths must satisfy the following constraint:

4 x + 2 π r = 2

Dividing by

2

, we obtain the simplified constraint:

2 x + π r = 1

This equation dictates that

x

and

r

are dependent; increasing the size of the square necessitates a decrease in the size of the circle.

The Area Function

Our goal is to minimize the total area

A

, defined as the sum of the area of the square and the area of the circle:

A = x^{2} + π r^{2}

To optimize this, we isolate

r

from the constraint equation:

r = \frac{1 - 2 x}{π}

Substituting this into the area function, we express

A

solely in terms of

x

A (x) = x^{2} + π (\frac{1 - 2 x}{π})^{2} = x^{2} + \frac{( 1 - 2 x ) ^{2}}{π}

The Calculus of Optimization

To find the minimum area, we calculate the derivative

\frac{d A}{d x}

and set it to zero.

Applying the power rule and the chain rule, we find:

\frac{d A}{d x} = 2 x + \frac{1}{π} \cdot 2 (1 - 2 x) \cdot (- 2)

Simplifying the derivative, we get:

\frac{d A}{d x} = 2 x - \frac{4 ( 1 - 2 x )}{π}

Setting the derivative to zero for optimization:

2 x - \frac{4 ( 1 - 2 x )}{π} = 0

The Elegant Conclusion

Multiplying the equation by

π

yields

2 π x - 4 + 8 x = 0

. Solving for

x

, we find:

x (2 π + 8) = 4 \Rightarrow x = \frac{2}{π + 4}

Substituting this value back into our expression for

r

r = \frac{1 - 2 ( \frac{2}{π + 4} )}{π} = \frac{1}{π + 4}

Observe that the side of the square

x

is exactly twice the radius

r

, meaning

x = 2 r

The total area is minimized when the side of the square is equal to the diameter of the circle.

Common Traps & FAQs

Wait, why do we use

4 x + 2 π r = 2

instead of just

x + r = 2

That is a classic trap! Remember,

x

is the side of the square and

r

is the radius of the circle. The wire forms the *boundary* of these shapes. The boundary of a square is its perimeter (

4 x

), and the boundary of a circle is its circumference (

2 π r

). You cannot add side lengths and radii directly because they represent different geometric properties. Always relate the wire length to the perimeter/circumference, not the dimensions themselves.

Is it always true that the side of the square equals the diameter of the circle for minimum area?

In this specific problem, yes! When you solve for

x

and

r

, you will find

x = \frac{2}{π + 4}

and

r = \frac{1}{π + 4}

. Since the diameter of the circle is

2 r

, we see that

x = 2 r

. This is a beautiful, elegant result that often appears in optimization problems involving geometric shapes. It suggests a certain 'balance' between the shapes that minimizes the total area.

How do I know if this critical point is a minimum or a maximum?

Great question! You can use the second derivative test. If you take the second derivative of the area function

A (x)

, you will find it is positive (

A^{''} (x) > 0

). A positive second derivative at a critical point confirms that the function is concave up, meaning the point is a local minimum. Since our function is a simple quadratic, this local minimum is also the global minimum.

Animated Solution for Mathematics - Differentiation: A wire of length $2$ units is cut into two parts which are bent respectively to form a square of side $= x$ units and a circle of radius $= r$ units. If the sum of the areas of the square and the circle so formed is minimum, then

Select Answer:

Visualized Solution

Visualizing the Wire Cut

Establishing the Perimeter Constraint

Simplifying the Constraint

Expressing $r$ in terms of $x$

The Total Area Function $A (x)$

Area in terms of a Single Variable

Condition for Minimum Area

Differentiating the Area Function

Setting Derivative to Zero

Solving for $x$

Finding the Radius $r$

Simplifying $r$

The Final Relation: $x = 2 r$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Constraint Equation

The Area Function

The Calculus of Optimization

The Elegant Conclusion

Similar Questions

A wire of length $36 m$ is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $k$ (meter), then $(\frac{4}{π} + 1) k$ is equal to .

A wire of length $20 m$ is to be cut into two pieces. A piece of length $ℓ_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $ℓ_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum then $(π ℓ_{1}) : ℓ_{2}$ is equal to:

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :

A wire of length $20 m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

Let a rectangle $A B C D$ of sides 2 and 4 be inscribed in another rectangle $P QR S$ such that the vertices of the rectangle $A B C D$ lie on the sides of the rectangle $P QR S$ . Let $a$ and $b$ be the sides of the rectangle $P QR S$ when its area is maximum. Then $(a + b)^{2}$ is equal to :

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:

A square piece of tin of side $30 cm$ is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in $cm^{2}$ ) is equal to

A box open from top is made from a rectangular sheet of dimension $a \times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to :

The circle $x^{2} + y^{2} = 1$ cuts the $x$ -axis at $P$ and $Q$ . Another circle with centre at $Q$ and variable radius intersects the first circle at $R$ above the $x$ -axis and the line segment $P Q$ at $S$ . Find the maximum area of the triangle $QS R$ .

A point $P$ is given on the circumference of a circle of radius $r$ . Chord $QR$ is parallel to the tangent at $P$ . Determine the maximum possible area of the triangle $P QR$ .

Select Answer:

Visualized Solution

Visualizing the Wire Cut

Establishing the Perimeter Constraint

Simplifying the Constraint

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The Total Area Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(x)

Area in terms of a Single Variable

Condition for Minimum Area

Differentiating the Area Function

Setting Derivative to Zero

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Finding the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

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The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Constraint Equation

The Area Function

The Calculus of Optimization

The Elegant Conclusion

Similar Questions

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the triangle PQR.

Expressing $r$ in terms of $x$

The Total Area Function $A (x)$

Solving for $x$

Finding the Radius $r$

Simplifying $r$

The Final Relation: $x = 2 r$

A point $P$ is given on the circumference of a circle of radius $r$ . Chord $QR$ is parallel to the tangent at $P$ . Determine the maximum possible area of the triangle $P QR$ .