Math Problem Statement

The diagram shows a square with side x cm and a circle with radius y cm. Write down an expression for the perimeter of the square and circle. The two shapes are made out of a piece of wire of total length 8 cm. Find an expression for x in terms of y. Show that the total area of the two shapes is given by the formula provided, and calculate the percentage of the wire used for the circle if the total area is minimized.

Solution

Let's break down the problem step by step from the image:

12a. Write an expression for the perimeter:

  • i. Of the square: The side of the square is given as xx cm.

    • The perimeter PsquareP_{\text{square}} of a square is 4×side4 \times \text{side}, so: Psquare=4xP_{\text{square}} = 4x
  • ii. Of the circle: The radius of the circle is given as yy cm.

    • The perimeter (circumference) PcircleP_{\text{circle}} of a circle is 2π×radius2\pi \times \text{radius}, so: Pcircle=2πyP_{\text{circle}} = 2\pi y

12b. The two shapes are made out of a piece of wire of total length 8 cm. Find an expression for xx in terms of yy.

The total length of the wire is 8 cm, so: 4x+2πy=84x + 2\pi y = 8 Now solve for xx in terms of yy: 4x=82πy4x = 8 - 2\pi y x=82πy4x = \frac{8 - 2\pi y}{4} Thus, the expression for xx is: x=2πy2x = 2 - \frac{\pi y}{2}

12c. Show that the total area of the two shapes is given by:

A=π4(π+4)y22πy+4A = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4

We need to find the total area of the square and the circle. The area of the square is x2x^2 and the area of the circle is πy2\pi y^2.

  1. Substitute the expression for xx from part (b) into x2x^2: x2=(2πy2)2x^2 = \left( 2 - \frac{\pi y}{2} \right)^2 Expanding: x2=42πy+π2y24x^2 = 4 - 2\pi y + \frac{\pi^2 y^2}{4}

  2. The total area is the sum of the area of the square and the area of the circle: A=(42πy+π2y24)+πy2A = \left( 4 - 2\pi y + \frac{\pi^2 y^2}{4} \right) + \pi y^2 Combine like terms: A=42πy+(π24+π)y2A = 4 - 2\pi y + \left( \frac{\pi^2}{4} + \pi \right) y^2 Factor the y2y^2 terms: A=42πy+π4(π+4)y2A = 4 - 2\pi y + \frac{\pi}{4}(\pi + 4) y^2 Thus, we have shown that the total area is: A=π4(π+4)y22πy+4A = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4

12d. If the total area of the two shapes is the smallest possible, what percentage of the wire is used for the circle?

To find this, we need to minimize the area function A(y)=π4(π+4)y22πy+4A(y) = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4. We will take the derivative of A(y)A(y) with respect to yy, set it to zero, and solve for yy.

Let me compute this.The critical value of yy that minimizes the total area is: y=4π+4y = \frac{4}{\pi + 4}

Next, we need to determine what percentage of the wire is used for the circle when y=4π+4y = \frac{4}{\pi + 4}.

  1. The length of wire used for the circle is the circumference 2πy2\pi y.
  2. The total length of wire is 8 cm.

The percentage of wire used for the circle is: Percentage=2πy8×100\text{Percentage} = \frac{2\pi y}{8} \times 100 Now, let's substitute the value of y=4π+4y = \frac{4}{\pi + 4} into this expression. I'll calculate this next.It seems like I can’t do more advanced data analysis right now. Please try again later.

You can manually compute the percentage using the following formula: Percentage=2π×4π+48×100\text{Percentage} = \frac{2\pi \times \frac{4}{\pi + 4}}{8} \times 100 This should give you the percentage of the wire used for the circle.

Do you have any questions or would you like help with any part of this problem?

Related Questions:

  1. How would you solve for xx in terms of yy if the total wire length changed?
  2. What happens to the total area if the square's side length and the circle's radius are doubled?
  3. How can you find the maximum area instead of the minimum?
  4. What percentage of the wire would be used for the square instead?
  5. How would the result change if the wire length was greater than 8 cm?

Tip: When solving optimization problems, always check for critical points using derivatives to ensure you're getting a minimum or maximum value.

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Math Problem Analysis

Mathematical Concepts

Perimeter
Area
Optimization
Quadratic Functions

Formulas

Perimeter of a square: P = 4x
Perimeter of a circle: P = 2πy
Total length of wire: 4x + 2πy = 8
Total area: A = x^2 + πy^2
Derivative to minimize total area

Theorems

Quadratic Expansion
Derivative for Optimization

Suitable Grade Level

Grade 10-12 (Precalculus)