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| Math Central | Quandaries & Queries |
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Question from yousef, a student: A man wishes to enclose two separate lots with 300m of fencing. One lot is a square and the other a rectangle whose length is twice its width. Find the dimensions of each lot if the total area is to be a minimum. |
Hi,
Suppose the square measures $X$ m on each side and the rectangle $W$ m wide and $L$ m long. Since the length of the rectangle is twice the width $L = 2 W.$ Hence the total area enclosed $A$ is given by
\[ A = x^2 + L \times W = X^2 + (2W) \times W = X^2 + 2W^2 \mbox{ square metres.}\]
The task is to use the calculus you know to maximize the expression for $A$ but $A$ is a function of two variables $X$ and $W.$ Use the fact that the length of fencing used is 300 m to express $A$ as a function of one variable and then use your calculus knowledge.
Write back if you need more assistance,
Penny
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