Question

(1)A window is in the form of rectangle surmounted by a semi-circle. The total perimeter of the window is
10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answer 1

Answer on Question #44598 – Math - Other

(1) A window is in the form of rectangle surmounted by a semi-circle. The total perimeter of the window is $10\,\mathrm{m}$. Find the dimensions of the window to admit maximum light through the whole opening.

Solution:

We start to solve with given data. We know that a window in the shape described above to have a maximum area (and hence let in the most light) and have a perimeter of $10\,\mathrm{m}$.

It also should be noted the reasonable assumption that the maximum amount of light would be admitted by the maximum area window.

Represent the sketch of the window below.

Let $y$ or $h$ represent the measure of the vertical dimension of the rectangular portion of the window. Let $x$ represent the horizontal dimension of the window. The value of $x$ is also the diameter of the semi-circular part of the window so we can write that $x = 2r$.

Then the perimeter of the window will be equal:

$$P(x, y) = 2y + x + \frac{\pi x}{2}$$

We know that the total perimeter of the window is equal to 10 meters. So, we can substitute the value of perimeter into the given formula:

$$2y + x + \frac{\pi x}{2} = 10$$

Now we can solve for $y$.

$$2y = 10 - x - \frac{\pi x}{2}$$

Divide both sides of the equation by 2 and simplify the equation.

$$y = 5 - \frac{x}{2} - \frac{\pi x}{4}$$ $$y = 5 - \frac {(\pi + 2) x}{4}$$

The area of the window is consisting of area semicircle and area of rectangle. So, we can write the formula for determination the area of the given window.

$$A(x, y) = x y + \frac {\pi \frac {x ^ {2}}{2}}{2}$$

Then we substitute $y$ to create a function for area in terms of $x$.

$$A(x, y) = x \left(5 - \frac {(\pi + 2) x}{4}\right) + \frac {\pi \frac {x ^ {2}}{2}}{2}$$

Simplify the equation.

$$A(x, y) = 5 x - \frac {(\pi + 2) x ^ {2}}{4} + \frac {\pi \frac {x ^ {2}}{2}}{2}$$

Finally simplify the value of Area.

$$A(x, y) = 5 x - \frac {(4 + \pi)}{8} x ^ {2}$$

Then we can apply two methods of solution. First implies the solution by consideration the quadratic equation or algebraically solve equation.

So, in our case we have a quadratic with a negative lead coefficient, hence the graph is a parabola that opens downward. That means the vertex is a maximum. It should also be noted that the vertex of $p(x) = ax^2 + bx + c$ is at $x_v = -\frac{b}{2a}$.

Applying this rule for our problem we can find the value of $x_v$.

$$x _ {m a x} = - \frac {b}{2 a} = \frac {- 5}{- 2 \left(\frac {(4 + \pi)}{8}\right)} = \frac {5}{\frac {(4 + \pi)}{4}} = \frac {2 0}{4 + \pi} \approx 2. 8 0 1 \mathrm {m e t e r}.$$

Now we can find the value of $y$. We have the formula for $y = 5 - \frac{(\pi + 2)x}{4}$, so we can substitute the value of $x$.

$$y = 5 - \frac {(\pi + 2)}{4} \frac {2 0}{(4 + \pi)}$$ $$y _ {m a x} = \frac {5 (\pi + 2)}{(4 + \pi)} \approx 3. 5 9 9 \mathrm {m e t e r}.$$

Then find the radius of the semi-circle is then:

$$\frac{x_{\max}}{2} = \frac{10}{4 + \pi} \approx 1.401 \text{ meter}.$$

Now we can consider another method of solution. We have a second degree polynomial equation, hence continuous and twice differentiable over the real numbers. Therefore there will be a local extremum at any point where the first derivative is equal to zero and that point will be a maximum if the second derivative is negative at that point.

$$\begin{array}{l} \frac{dA}{dx} = 5 - \frac{(4 + \pi)}{4}x \\ \frac{dA}{dx} = 0 \\ 5 - \frac{(4 + \pi)}{4}x = 0 \\ \end{array}$$

Simplify by solving for $x$.

$$- \frac{(4 + \pi)}{4}x = -5$$

Multiply both sides of the equation by -4.

$$(4 + \pi)x = 20$$

The value of $x$ will be equal.

$$x = \frac{20}{4 + \pi}$$

Then we find the second derivative.

$$\frac{d^2A}{dx^2} = - \frac{(4 + \pi)}{4} < 0$$

Hence value of $x$ is equal to $x = \frac{20}{4 + \pi}$ and the coordinate of the maximum point.

The $y$ dimension and the semi-circle radius are calculated as above.

**Answer**: The dimensions of the window to admit maximum light through the whole opening will be the horizontal dimension of the window $x_{\max} = \frac{20}{4 + \pi} \approx 2.801$ meter, radius of the semi-circle $\frac{x_{\max}}{2} = \frac{10}{4 + \pi} \approx 1.401$ meter and the measure of the vertical dimension of the rectangular portion $y_{\max} = \frac{5(\pi + 2)}{(4 + \pi)} \approx 3.599$ meter.

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