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

3 Answers
Feb 5, 2017

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

Explanation:

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.

Feb 6, 2017

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

Explanation:

Now using the Lagrange Multipliers technique.

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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#

Feb 6, 2017

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

Explanation:

And now with rotations.

enter image source here

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#