Problem 45 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Lamar is an artist who has been commissioned to construct an ornate window. The window is to be in the form of an equilateral triangle surmounted on a rectangle, and the entire window is to have a perimeter of 20 feet. The rectangular part will be made of clear glass and will admit twice as much light as the stained glass triangular part. What dimensions should Lamar choose for the window if he wants it to admit the maximum amount of light?

Step-by-Step Solution

Answer

Lamar should choose the dimensions approximately \( w = 2.35 \text{ feet} \) and \( h = 5.3 \text{ feet} \).

Step 1: Understand the problem

The window is made of two parts: a triangle on top of a rectangle. We need to maximize the amount of light admitted, considering the clear glass (rectangle) admits twice as much light as the stained glass (triangle). The total perimeter is 20 feet.

Step 2: Define variables

Let the width of the rectangle be denoted by \( w \) and the height of the rectangle by \( h \). Since the triangle is equilateral, each side of the triangle is also \( w \).

Step 3: Express the perimeter in terms of variables

The perimeter of the window is made up of the three sides of the triangle plus twice the height of the rectangle and its width. The perimeter equation is:\[ P = w + 3w + 2h = 20 \] Simplify to find:\[ 4w + 2h = 20 \] or alternatively, solving for one variable:\[ 2w + h = 10 \] so, \( h = 10 - 2w \).

Step 4: Write the light admission function

The light admitted by the rectangle (clear glass) and triangle (stained glass) needs to be taken into account. Let the light admitted by the stained glass triangle be \( L_T \) and the clear glass rectangle be \( L_R \).Since the rectangle admits twice as much light:\[ L_R = 2wh \]The area of the triangle is \( \frac{\root3 \times w^2}{4} \), hence light admitted by the triangle is:\[ L_T = \frac{\root3 \times w^2}{4} \]The total light admitted is:\[ L_{total} = 2wh + \frac{\root3 \times w^2}{4} \]

Step 5: Substitute \( h \) and form the light function

Using \( h = 10 - 2w \), the light function becomes:\[ L_{total} = 2w(10 - 2w) + \frac{\root3 \times w^2}{4} \]Simplify to:\[ L_{total} = 20w - 4w^2 + \frac{\root3 \times w^2}{4} \]

Step 6: Differentiate and find the critical points

To find the maximum, take the derivative of the light function with respect to \( w \) and set it to zero:\[ \frac{dL_{total}}{dw} = 0 \ \frac{d}{dw} \big(20w - 4w^2 + \frac{\root3 \times w^2}{4}\big) = 20 - 8w + \frac{\root3 \times w}{2} = 0 \]Solve for \( w \) to find critical points.

Step 7: Solve derivative equation for critical points

Rearrange to solve for \( w \):\[ 20 - 8w + \frac{\root3 \times w}{2} = 0 \ 40 - 16w + \root3 \times w = 0 \ w = \frac{40}{16 - \root 3} \ \approx 2.35 \]

Step 8: Determine corresponding h value

Substitute \( w \) back into \( h = 10 - 2w \):\[ h = 10 - 2 \times 2.35 \approx 5.3 \]

Step 9: Conclusion

The dimensions Lamar should choose to maximize light admitted are approximately \( w = 2.35 \text{ feet} \ h = 5.3 \text{ feet} \).

Key Concepts

Perimeter Optimization

Perimeter optimization is a common problem in calculus where you seek to maximize or minimize a quantity subject to a given perimeter. In this problem, Lamar's window has a **20-foot perimeter**, composed of an equilateral triangle on top of a rectangle. This total length constraint is crucial because it must always hold true. We broke down the constraints into the equation \(4w + 2h = 20\), solving for variable combinations that sum to **20 feet**. Any modifications in width \(w\) and height \(h\) need to satisfy this perimeter constraint while optimizing for other quantities, in this case, maximizing light admission.

Light Admission Function

Admittance of light is integral to the problem. Clear glass in the rectangle admits twice as much light compared to stained glass in the triangle. This leads us to set up a function for total light using the areas of the triangle and rectangle. \( L_{total} = 2wh + \frac{\root3 w^2}{4} \) gives us the total light admitted. Here, \( 2wh \) represents light from the rectangle while the triangular section's light is affected by its relatively lower stained glass light admission. By substituting \( h = 10 - 2w \) into this function, we translate the light function solely in terms of \( w \), making it easier to differentiate and optimize.

Critical Points in Calculus

Critical points allow us to find where functions reach their maximum or minimum values, which for Lamar's problem would mean **maximized light**. By differentiating the total light function with respect to the width \( w \) and setting the derivative to zero, we find \( w \)'s critical points. The derivative, \( \frac{dL_{total}}{dw} \), simplifies to \( 20 - 8w + \frac{\root3 w}{2} \). Solving \( 20 - 8w + \frac{\root3 w}{2} = 0 \) gives \( w \) approximately \( 2.35 \) feet. The corresponding height can be calculated as \( h = 10 - 2(2.35) \), giving \( h \) as roughly \( 5.3 \) feet. These dimensions ensure optimal light admittance under the given constraints.