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

Answer 1

#1/2(L+W)^2#

Let us set up a concrete scenario...

Start with a rectangle with vertices:

#(L/2, W/2)#
#(-L/2, W/2)#
#(-L/2, -W/2)#
#(L/2, -W/2)#
Rotate it anticlockwise about the origin by #theta# to give a rectangle with vertices:
#(L/2 cos theta - W/2 sin theta, W/2 cos theta + L/2 sin theta)#
#(-L/2 cos theta - W/2 sin theta, W/2 cos theta - L/2 sin theta)#
#(-L/2 cos theta + W/2 sin theta, -W/2 cos theta - L/2 sin theta)#
#(L/2 cos theta + W/2 sin theta, -W/2 cos theta + L/2 sin theta)#
(For simplicity, just consider #0 <= theta <= pi/2# so we don't have to be concerned about a variety of cases, etc.)
Then the circumscribing rectangle with sides parallel to the #x# and #y# axes has area:
#(L cos theta + W sin theta)(W cos theta + L sin theta) = (L^2+W^2)cos theta sin theta+WL#
#color(white)((L cos theta + W sin theta)(W cos theta + L sin theta)) = 1/2(L^2+W^2)sin 2theta+WL#
This takes its maximum value when #sin 2theta = 1#, e.g. when #theta = pi/4#

So the maximum area is:

#WL+1/2(L^2+W^2) = 1/2(L+W)^2#

Unsurprisingly, this is when the circumscribing rectangle is a square.

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Answer 2

#1/2(L+W)^2#

And now with rotations.

The circumscribed quadrilateral area is given by

#A(theta)=2(1/2(Wsintheta)(Wcostheta)+1/2(Lsintheta)(Lcostheta))+LW#

so

#A(theta)=(W^2+L^2)sintheta costheta+LW#

Now the maximum is at the solution of

#(dA)/(d theta) = (W^2+L^2)(1-2sin^2theta)=0#

giving #theta={-3pi/4,-pi/4,pi/4,3pi/4}#. Between those solutions we will choose the maximum. The local maxima are located at points in which #(d^2A)/(d theta^2) < 0#

so the solution is for #theta=pi/4# or #theta=-3pi/4#

#A(pi/4)=(W^2+L^2)(1/sqrt(2))^2+LW=1/2(L+W)^2#

because at this point

#(d^2A)/(d theta^2)=-4(W^2+L^2)costheta sintheta=-2(W^2+L^2)<0#

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Answer 3

#1/2(L+ W)^2#

Now using the Lagrange Multipliers technique.

The circumscribed rectangle has the side dimensions

#(a+b)# and #(c+d)# so the sough area is #(a+b)(c+d)#

The restrictions are

#a^2+b^2=L^2# and
#c^2+d^2=W^2#

The lagrangian is

#Phi(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)#

The stationary points are the solutions of

#grad Phi=(Phi_a,Phi_b,Phi_c,Phi_d,Phi_(lambda_1),Phi_(lambda_2))=0#

or

#{(c + d + 2 a lambda_1=0),(c + d + 2 blambda_2=0),(a + b + 2 c lambda_2=0),(a + b + 2 d lambda_1=0),(a^2 + d^2 - L^2=0),(b^2 + c^2 - W^2=0):}#

Solving for #a,b,c,d,lambda_1,lambda_2# we get at

#((a=L/sqrt[2]),(b=W/sqrt[2]),(c=W/sqrt[2]),(d=L/sqrt[2]),(lambda_1=-(L +W)/(2 L)),(lambda_1=-(L +W)/(2 W)))#

and the maximum area is

#(a+b)(c+d)=1/2(L+ W)^2#

Of course the minimum area circumscribing rectangle has the area #L cdot W#

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Answer 4

The maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W is achieved when the circumscribing rectangle has its vertices touching the midpoints of the sides of the given rectangle. In this scenario, the length of the circumscribing rectangle is equal to twice the length of the given rectangle (2L), and the width of the circumscribing rectangle is equal to twice the width of the given rectangle (2W). Therefore, the maximum area ( A_{\text{max}} ) of the circumscribing rectangle is given by the formula:

[ A_{\text{max}} = (2L) \times (2W) = 4LW ]

So, the maximum area of the circumscribing rectangle is four times the area of the given rectangle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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