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# Geometry

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There are many ways to circumscribe a rectangle R about a 5 x 10 rectangle so that each vertex of the 5 x 10 rectangle is on a different side of R. Rectangle R's area is 110. what is the maximum area of a rectangle R that can be circumscribed about a 5 x 10 rectangle?

Mar 15, 2019

#1
+23835
+3

There are many ways to circumscribe a rectangle R about a 5 x 10 rectangle so
that each vertex of the 5 x 10 rectangle is on a different side of R.
Rectangle R's area is 110. what is the maximum area of a rectangle R
that can be circumscribed about a 5 x 10 rectangle?

Let:

$$\begin{array}{|rcll|} \hline a &=& W\sin(\theta) \\ c &=& W\cos(\theta) \\ \hline \end{array} \begin{array}{|rcll|} \hline d &=& L\sin(\theta) \\ b &=& L\cos(\theta) \\ \hline \end{array}$$

The area A of the circumscribing rectangle:

$$\begin{array}{|rcll|} \hline \mathbf{A(\theta)} &\mathbf{=}& \mathbf{(a+b)(c+d)} \\\\ A(\theta) &=& \Big( W\sin(\theta)+L\cos(\theta) \Big) \Big( W\cos(\theta)+L\sin(\theta) \Big) \\ &=& W^2\sin(\theta)\cos(\theta) +L\cdot W\sin^2(\theta) + LW\cos^2(\theta) + L^2\sin(\theta)\cos(\theta) \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W\Big((\sin^2(\theta) + \cos^2(\theta)\Big) \quad | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W \quad | \quad \sin(\theta)\cos(\theta) = \dfrac{\sin(2\theta)}{2} \\ &=& (W^2+L^2)\dfrac{\sin(2\theta)}{2} +L\cdot W \\ && \color{red}\underline{\text{maximize}}:\\ && \qquad \color{red}\sin(2\theta) = 1 \\ && \qquad \color{red}2\theta = \arcsin(1) \\ && \qquad \color{red} 2\theta = 90^\circ \\ && \qquad \color{red} \theta = 45^\circ \\ A_{\text{max}}(45^\circ) &=& (W^2+L^2)\dfrac{\sin(2\cdot 45^\circ)}{2} +L\cdot W \\ &=& (W^2+L^2)\dfrac{\sin(90^\circ)}{2} +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2) +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2+2LW) \\ &=& \dfrac{1}{2} (W+L)^2 \\ \hline \end{array}$$

The maximum area:

$$\begin{array}{|rcll|} \hline A_{\text{max}} &=& \dfrac{1}{2} (W+L)^2 \\\\ &=& \dfrac{1}{2} (5+10)^2 \\\\ &=& \dfrac{15^2}{2} \\\\ &=& \dfrac{225}{2} \\\\ &=& 112.5 \\ \hline \end{array}$$

The maximum area is 112.5

Mar 15, 2019
edited by heureka  Mar 15, 2019
edited by heureka  Mar 15, 2019

#1
+23835
+3

There are many ways to circumscribe a rectangle R about a 5 x 10 rectangle so
that each vertex of the 5 x 10 rectangle is on a different side of R.
Rectangle R's area is 110. what is the maximum area of a rectangle R
that can be circumscribed about a 5 x 10 rectangle?

Let:

$$\begin{array}{|rcll|} \hline a &=& W\sin(\theta) \\ c &=& W\cos(\theta) \\ \hline \end{array} \begin{array}{|rcll|} \hline d &=& L\sin(\theta) \\ b &=& L\cos(\theta) \\ \hline \end{array}$$

The area A of the circumscribing rectangle:

$$\begin{array}{|rcll|} \hline \mathbf{A(\theta)} &\mathbf{=}& \mathbf{(a+b)(c+d)} \\\\ A(\theta) &=& \Big( W\sin(\theta)+L\cos(\theta) \Big) \Big( W\cos(\theta)+L\sin(\theta) \Big) \\ &=& W^2\sin(\theta)\cos(\theta) +L\cdot W\sin^2(\theta) + LW\cos^2(\theta) + L^2\sin(\theta)\cos(\theta) \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W\Big((\sin^2(\theta) + \cos^2(\theta)\Big) \quad | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \\ &=& (W^2+L^2)\sin(\theta)\cos(\theta) +L\cdot W \quad | \quad \sin(\theta)\cos(\theta) = \dfrac{\sin(2\theta)}{2} \\ &=& (W^2+L^2)\dfrac{\sin(2\theta)}{2} +L\cdot W \\ && \color{red}\underline{\text{maximize}}:\\ && \qquad \color{red}\sin(2\theta) = 1 \\ && \qquad \color{red}2\theta = \arcsin(1) \\ && \qquad \color{red} 2\theta = 90^\circ \\ && \qquad \color{red} \theta = 45^\circ \\ A_{\text{max}}(45^\circ) &=& (W^2+L^2)\dfrac{\sin(2\cdot 45^\circ)}{2} +L\cdot W \\ &=& (W^2+L^2)\dfrac{\sin(90^\circ)}{2} +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2) +L\cdot W \quad | \quad \sin(90^\circ) = 1 \\ &=& \dfrac{1}{2} (W^2+L^2+2LW) \\ &=& \dfrac{1}{2} (W+L)^2 \\ \hline \end{array}$$

The maximum area:

$$\begin{array}{|rcll|} \hline A_{\text{max}} &=& \dfrac{1}{2} (W+L)^2 \\\\ &=& \dfrac{1}{2} (5+10)^2 \\\\ &=& \dfrac{15^2}{2} \\\\ &=& \dfrac{225}{2} \\\\ &=& 112.5 \\ \hline \end{array}$$

The maximum area is 112.5

heureka Mar 15, 2019
edited by heureka  Mar 15, 2019
edited by heureka  Mar 15, 2019
#2
+106532
0

Nice, heureka !!!!

CPhill  Mar 15, 2019
#3
+23835
+2

Thank you, CPhill.

heureka  Mar 15, 2019