Question

What is the maximum area of a rectangle that can be circumscribed about a given rectangle of length L and width W?

Answer 1

We are given a rectangle of length L and width W. We are asked to find the area of a rectangle that can be circumscribed on this given rectangle of given dimensions. We will first draw the corresponding diagram to get a clear picture. We will then name each of the sides on the basis of the rectangle intersecting the external rectangle. We will use the Lagrange Multipliers technique to find the dimensions of the new rectangle circumscribed around the given rectangle of the given dimensions. Then, having the dimensions, we will find the area of the rectangle. Hence, we will have the required area of the rectangle.

Complete step-by-step answer:
According to the given question, we are given a rectangle having dimensions of length as L and width as W. Then, we circumscribed another rectangle around the given rectangle and we are asked to find the area of this new rectangle formed.
Firstly, we will draw an appropriate diagram for the given question and we have something like,

Here, we have named the sides on the basis of the intersection of the vertices of the rectangle inside with the circumscribed rectangle.
In order to solve for the area of this circumscribed rectangle in terms of the given lengths L and W, we will use the Lagrange Multipliers technique.
As per Lagrange Multipliers, if $L$ is the Lagrangian function over the integer ‘x’ and Lagrange multiplier $\lambda$, we get,
$L(x,\lambda )=f(x)-\lambda g(x)$
where $f(x)$ is the function and $g(x)$ is the equality constraint.
From the figure, we have the dimensions of the circumscribed rectangle as,
$(a+b)$ and $(c+d)$ and so the area comes out to be $(a+b)(c+d)$.
We have the restrictions as,
${{a}^{2}}+{{b}^{2}}={{L}^{2}}$
${{c}^{2}}+{{d}^{2}}={{W}^{2}}$
Now, we will apply the Lagrangian are,
$L(a,b,c,d,{{\lambda }_{1}},{{\lambda }_{2}})=(a+b)(c+d)+{{\lambda }_{1}}({{a}^{2}}+{{b}^{2}}-{{L}^{2}})+{{\lambda }_{2}}({{c}^{2}}+{{d}^{2}}-{{W}^{2}})$
Solving the above expression further, we get,
$\Rightarrow L(a,b,c,d,{{\lambda }_{1}},{{\lambda }_{2}})=ac+ad+bc+bd+{{\lambda }_{1}}({{a}^{2}}+{{b}^{2}}-{{L}^{2}})+{{\lambda }_{2}}({{c}^{2}}+{{d}^{2}}-{{W}^{2}})$-----(1)
We will now take partial derivatives of each of the variables as well as the lagrange multipliers, we get,
$\nabla L=({{L}_{a}},{{L}_{b}},{{L}_{c}},{{L}_{d}},{{L}_{{{\lambda }_{1}}}},{{L}_{{{\lambda }_{2}}}})=0$
Taking the partial derivatives of equation (1) w.r.t ‘a’, we get,
$c+d+2a{{\lambda }_{1}}=0$-----(2)
Next, taking the partial derivative w.r.t ‘b’, we get,
$c+d+2b{{\lambda }_{1}}=0$-----(3)
Next, taking the partial derivative w.r.t ‘c’, we get,
$a+b+2c{{\lambda }_{2}}=0$-----(4)
Next, taking the partial derivative w.r.t ‘d’, we get,
$a+b+2d{{\lambda }_{2}}=0$-----(5)
We will now take the partial derivative of the lagrange multiplier ${{\lambda }_{1}}$, we have,
${{a}^{2}}+{{b}^{2}}-{{L}^{2}}=0$-----(6)
Taking the partial derivative of the lagrange multiplier ${{\lambda }_{2}}$, we have,
${{c}^{2}}+{{d}^{2}}-{{W}^{2}}=0$------(7)
We will now solve the above equations to find the values of a, b, c and d. and we have,
Subtracting equation (2) and equation (3), we get,
$2a{{\lambda }_{1}}-2b{{\lambda }_{1}}=0$
We get the value as, ${{\lambda }_{1}}=0$
Similarly, on subtracting the equation (4) and (5), we get,
${{\lambda }_{2}}=0$
Applying these on other equations, we get the value of,
$a=\dfrac{L}{\sqrt{2}}$
$b=\dfrac{W}{\sqrt{2}}$
$c=\dfrac{W}{\sqrt{2}}$
and $d=\dfrac{L}{\sqrt{2}}$
Now, substituting these values in the formula of area of the rectangle, we get,
$(a+b)(c+d)$
$\Rightarrow \left( \dfrac{L}{\sqrt{2}}+\dfrac{W}{\sqrt{2}} \right)\left( \dfrac{W}{\sqrt{2}}+\dfrac{L}{\sqrt{2}} \right)$
$\Rightarrow \left( \dfrac{L+W}{\sqrt{2}} \right)\left( \dfrac{L+W}{\sqrt{2}} \right)$
$\Rightarrow \dfrac{1}{2}{{\left( L+W \right)}^{2}}$
Therefore, the area of the circumscribed rectangle is $\dfrac{1}{2}{{\left( L+W \right)}^{2}}$

Note: The above solution can also be done using the polar coordinates as well as using the trigonometric functions. While carrying out the above solution, the Lagrange multiplier should be correctly known and should be carried out step wise to avoid missing any terms. Also, the partial derivatives done should be correct else the whole answer can get wrong too after coming this far.