A unique window has the shape of a rectangle with a quarter circle on top. What height, width, and radius
will maximize the enclosed area if the outside perimeter is 4 m?
Assuming the radius of the 1/4 circle is the width of the rectangle, then if the rectangle has height h, the perimeter is
p = π/2 r + r + h + r + h = 2h + (2 + π/2)r = 4
so h = 4 - (2 + π/2)r
The area is
a = hr + 1/4 πr^2
= (4 - (2 + π/2)r)*r + 1/4 πr^2
= 4r - (2 + π/4) r^2
da/dr = 4 - (4 + π/2)r
da/dr=0 when r = 8/(π+8)
now just find h. The width is the same as the radius r.
Well, I guess you could say this window is a real shape-shifter! Let's solve this mathematical conundrum while having some fun.
Let's assume that the rectangle's height is "h", its width is "w", and the radius of the quarter circle on top is "r". To maximize the enclosed area, we need to find the values of h, w, and r that satisfy the outside perimeter of 4 m.
Considering this, we have the following equation for the perimeter:
2w + πr + 2h = 4
To make it easier, let's take a break from the numbers and dive into visuals. Imagine a rectangle sitting on a quarter circle, feeling like it's on top of the world!
Now, since we know that the quarter circle sits on top of the rectangle, its diameter will be equal to the width of the rectangle. This means that:
2r = w
We can substitute this expression into our original perimeter equation, giving us:
2(2r) + πr + 2h = 4
Simplifying further, we get:
4r + πr + 2h = 4
Now, it's time for a clever move to maximize the fun! Since we want to maximize the area, we can transform this problem into an area-maximizing challenge. The enclosed area will be the sum of the rectangle's area and the quarter circle's area. Let's name this mind-boggling area "A," because we're all about short and sweet.
The area of the rectangle is simply given by:
Rectangle Area = w * h
Plugging in our expression for w in terms of r, we get:
Rectangle Area = 2r * h
Now, let's focus on the quarter circle's area, which is:
Quarter Circle Area = (π * r^2) / 4
Next-level math calls for taking the derivative of the area function with respect to r and setting it equal to zero to find the maximum:
dA/dr = 2h - (π * r) / 2 = 0
Simplifying this equation gives us:
h = πr / 4
Now that we've found the relationship between h and r, let's substitute it back into our perimeter equation:
4r + πr + 2(πr/4) = 4
Simplifying this equation further:
4r + πr + 0.5πr = 4
Combining like terms and dividing everything by 4:
4.5r + πr = 4
Now we're left with an equation for the radius r, which should bring us to the grand finale.
r(4.5 + π) = 4
r = 4 / (4.5 + π)
After some quick calculations:
r ≈ 0.542 m
Now that we know the radius, we can find the width of the rectangle (w) using our earlier equation:
w = 2r
w ≈ 1.084 m
To find the height of the rectangle (h), we can use the relationship we discovered earlier:
h = πr / 4
h ≈ π * 0.542 / 4
h ≈ 0.426 m
So, the height is approximately 0.426 m, the width is about 1.084 m, and the radius we've all been waiting for is around 0.542 m. Voilà! We've successfully cracked the code and maximized the enclosed area of this unique window.
To find the dimensions that maximize the enclosed area of the unique window, we need to maximize the area function. Let's break down the problem step by step:
Step 1: Define the variables:
Let's define the height of the rectangle as "h", the width of the rectangle as "w", and the radius of the quarter circle as "r".
Step 2: Write the equations:
The outside perimeter of the window is given as 4 m. This can be expressed as an equation:
P = 2h + w + πr/2 = 4
Step 3: Rewrite the equation:
From the equation in Step 2, we can isolate one variable. Let's isolate w:
w = 4 - 2h - πr/2
Step 4: Write the area function:
The area enclosed by the window is the sum of the rectangle's area and the quarter circle's area. Let's define the area function:
A = h * w + (πr^2)/4
Step 5: Substitute the equation for w into the area function:
A = h * (4 - 2h - πr/2) + (πr^2)/4
Step 6: Simplify the area function:
A = 4h - 2h^2 - πhr/2 + (πr^2)/4
Step 7: Find the derivative of the area function with respect to one variable:
To find the dimensions that maximize the area, we need to find the critical points of the area function. For that, we'll take the derivative of the area function with respect to one variable. Let's find the partial derivatives with respect to h and r separately:
∂A/∂h = 4 - 4h - πr/2
∂A/∂r = - (πh/2) + (πr/2)
Step 8: Set the derivatives equal to zero and solve for the variables:
To find the critical points, we'll set the partial derivatives equal to zero and solve for h and r:
For ∂A/∂h:
4 - 4h - πr/2 = 0
-4h = -4 + πr/2
h = 1 - (πr/8) ----(Equation 1)
For ∂A/∂r:
- (πh/2) + (πr/2) = 0
-πh/2 + πr/2 = 0
πr/2 = πh/2
r = h -----(Equation 2)
Step 9: Substitute the value of "r" from Equation 2 into Equation 1:
h = 1 - (πr/8)
h = 1 - (πh/8)
h + (πh/8) = 1
(8h + πh) / 8 = 1
8h + πh = 8
h(8 + π) = 8
h = 8 / (8 + π)
Step 10: Substitute the value of "h" into Equation 2 to find "r":
r = h
r = 8 / (8 + π)
Step 11: Substitute the values of "h" and "r" into the equation for "w" from Step 3:
w = 4 - 2h - πr/2
w = 4 - 2(8/(8 + π)) - π(8/(8 + π))/2
Simplifying the equation will give you the value of "w".
Therefore, the dimensions that maximize the enclosed area of the unique window are:
- Height (h): 8 / (8 + π) meters
- Width (w): Calculated using the equation in Step 11
- Radius (r): 8 / (8 + π) meters
To solve this problem, we need to maximize the area of the window while keeping the outside perimeter fixed at 4 meters.
Let's analyze the problem step by step:
1. Identify the variables:
- Width of the rectangle (w)
- Height of the rectangle (h)
- Radius of the quarter circle (r)
2. Set up the equations:
- Perimeter equation: 2w + h + πr + 2r = 4
3. Simplify the perimeter equation to solve for one variable in terms of the other variables:
- h = 4 - 2w - πr - 2r
4. Determine the area equation of the window:
- Area = Area of rectangle + Area of quarter circle
- Area = w * h + (1/4)πr^2
5. Substitute h from the perimeter equation into the area equation:
- Area = w * (4 - 2w - πr - 2r) + (1/4)πr^2
6. Simplify the equation:
- Area = 4w - 2w^2 - πwr - 2wr + (1/4)πr^2
7. To maximize the area, we need to differentiate the equation with respect to each variable (w and r) and set the derivatives equal to zero:
- d(Area)/dw = 4 - 4w - πr = 0
- d(Area)/dr = -πw - 2πr + (1/2)πr = 0
8. Solve the system of equations:
- From the first equation: 4 - 4w - πr = 0 => 4w + πr = 4
- From the second equation: -πw - 2πr + (1/2)πr = 0 => πw - (3/2)πr = 0
9. Solve the system of equations to find the values of w and r:
- Multiply the second equation by 2: 2πw - 3πr = 0
- Add this equation to the first equation: 4w + πr + 2πw - 3πr = 4
- Simplify and solve: 6w - 2πr = 4 => 6w = 2πr + 4 => w = (2πr + 4)/6
- Substitute w back into the first equation: 4(2πr + 4)/6 + πr = 4
- Solve for r.
10. Once you find the value of r, substitute it back into the expression for w, and then calculate the corresponding value of h.
Therefore, to maximize the area of the unique window, you need to solve the system of equations obtained in step 9 to find the values of the width (w), height (h), and radius (r) that will give you the maximum enclosed area.