
A window is in the shape of a rectangle surmounted by a semicircle opening. The total perimeter of the window is $10m$ . Find the dimension of the window to admit maximum through the whole opening.
Answer
585.9k+ views
Hint: The word surmounted means it is on top of a rectangular window that has a semi-circle top. If a function is double differentiated then it can give you a maximum or minimum point and the perimeter of the circle is $(\pi r)$ , whereas that of the rectangle is $(length + base)$ .
Complete step-by-step solution:
Given: Perimeter of the window is $10m$ .
Let the radius of the semi-circle, length and breadth of the rectangle be $r,x$ and $y$ respectively.
Therefore, from the figure we can determine that
$AE = r$
$AB = x = 2r$ , since the semi-circle is mounted over the rectangle ---- (1)
$AD = y$
Now, according to the question,
Perimeter of the window is $10m$.
$
\Rightarrow x + 2y + \pi r = 10 \\
\Rightarrow (2r + \pi r) - 10 = - 2y \\
\Rightarrow 2y = 10 - (\pi + 2)r \\
$
$y = \dfrac{{\left( {10 - (\pi + 2)r} \right)}}{2}$ -------- (2)
To admit the maximum amount of light, the area of the window should be maximum. Assuming the area of the window as, area of the rectangle $+$ area of the semicircle.
$
A = xy + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = (2r)\left( {\dfrac{{10 - (\pi + 2)r}}{2}} \right) + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = 10r - \pi {r^2} - 2{r^2} + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = 10r - 2{r^2} - \dfrac{{\pi {r^2}}}{2} \\
$
Condition for maxima and minima is to differentiate the area,
$
\dfrac{{dA}}{{dr}} = 0 \\
\Rightarrow 10 - 4r - \pi r = 0 \\
\Rightarrow r = \dfrac{{10}}{{4 + \pi }} \\
$
Now, double differentiating the area, we have
$\dfrac{{{d^2}A}}{{d{r^2}}} = - 4 - \pi < 0$
For the $r = \dfrac{{10}}{{(4 + \pi )}}$ , $A$ will be maximum here $A$ stands for area.
Length of the rectangular part $ = \dfrac{{20}}{{(4 + \pi )}}m$ [from equation (1)]
Breadth of the rectangular part \[ = \dfrac{{10 - (\pi + 2)r}}{2}m\] [from equation (2)]
$
\Rightarrow y = \dfrac{{10 - \dfrac{{(\pi + 2)10}}{{4 + \pi }}}}{2} \\
\Rightarrow y = \dfrac{{10}}{{4 + \pi }} \\
$
Thus, the dimension of the window which will admit the maximum amount of light is
$x = \dfrac{{20}}{{4 + \pi }}m$ and $y = \dfrac{{10}}{{4 + \pi }}m$
Note: In this type of questions students often make the mistake that they do not double differentiate and take the area of the rectangle part as $2(x + y)$ which is incorrect. Do not make this mistake.
Complete step-by-step solution:
Given: Perimeter of the window is $10m$ .
Let the radius of the semi-circle, length and breadth of the rectangle be $r,x$ and $y$ respectively.
Therefore, from the figure we can determine that
$AE = r$
$AB = x = 2r$ , since the semi-circle is mounted over the rectangle ---- (1)
$AD = y$
Now, according to the question,
Perimeter of the window is $10m$.
$
\Rightarrow x + 2y + \pi r = 10 \\
\Rightarrow (2r + \pi r) - 10 = - 2y \\
\Rightarrow 2y = 10 - (\pi + 2)r \\
$
$y = \dfrac{{\left( {10 - (\pi + 2)r} \right)}}{2}$ -------- (2)
To admit the maximum amount of light, the area of the window should be maximum. Assuming the area of the window as, area of the rectangle $+$ area of the semicircle.
$
A = xy + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = (2r)\left( {\dfrac{{10 - (\pi + 2)r}}{2}} \right) + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = 10r - \pi {r^2} - 2{r^2} + \dfrac{{\pi {r^2}}}{2} \\
\Rightarrow A = 10r - 2{r^2} - \dfrac{{\pi {r^2}}}{2} \\
$
Condition for maxima and minima is to differentiate the area,
$
\dfrac{{dA}}{{dr}} = 0 \\
\Rightarrow 10 - 4r - \pi r = 0 \\
\Rightarrow r = \dfrac{{10}}{{4 + \pi }} \\
$
Now, double differentiating the area, we have
$\dfrac{{{d^2}A}}{{d{r^2}}} = - 4 - \pi < 0$
For the $r = \dfrac{{10}}{{(4 + \pi )}}$ , $A$ will be maximum here $A$ stands for area.
Length of the rectangular part $ = \dfrac{{20}}{{(4 + \pi )}}m$ [from equation (1)]
Breadth of the rectangular part \[ = \dfrac{{10 - (\pi + 2)r}}{2}m\] [from equation (2)]
$
\Rightarrow y = \dfrac{{10 - \dfrac{{(\pi + 2)10}}{{4 + \pi }}}}{2} \\
\Rightarrow y = \dfrac{{10}}{{4 + \pi }} \\
$
Thus, the dimension of the window which will admit the maximum amount of light is
$x = \dfrac{{20}}{{4 + \pi }}m$ and $y = \dfrac{{10}}{{4 + \pi }}m$
Note: In this type of questions students often make the mistake that they do not double differentiate and take the area of the rectangle part as $2(x + y)$ which is incorrect. Do not make this mistake.
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