AMC8 2008

AMC8 2008 · Q17

AMC8 2008 · Q17. It mainly tests Linear equations, Area & perimeter.

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

奥斯本女士让班上的每个学生画一个周长为 $50$ 单位的整数边长矩形。所有学生都计算了自己画的矩形的面积。最大可能面积与最小可能面积的差是多少？

(A) 76 76

(B) 120 120

(D) 132 132

(E) 136 136

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The formula for the perimeter of a rectangle is $2l+2w$, so $2l+2w=50$, and $l+w=25$. Make a chart of the possible widths, lengths, and areas, assuming all the widths are shorter than all the lengths. $\begin{array}{|c|cccccccccccc|} \hline \text{Width} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{Length} & 24 & 23 & 22 & 21 & 20 & 19 & 18 & 17 & 16 & 15 & 14 & 13 \\ \hline \text{Area} & 24 & 46 & 66 & 84 & 100 & 114 & 126 & 136 & 144 & 150 & 154 & 156 \\ \hline \end{array}$ The largest possible area is $13\times 12=156$ and the smallest is $1\times 24=24$, for a difference of $156-24=132$ square units. Note: The product of two numbers with a fixed sum increases as the numbers get closer together. That means, given the same perimeter, the square has a larger area than any rectangle, and a rectangle with a shape closest to a square will have a larger area than other rectangles with equal perimeters.

答案（D）：长方形周长的公式是 $2l+2w$，所以 $2l+2w=50$，并且 $l+w=25$。在假设所有宽都小于所有长的情况下，制作一个可能的宽、长与面积的表格。 $\begin{array}{|c|cccccccccccc|} \hline \text{宽} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{长} & 24 & 23 & 22 & 21 & 20 & 19 & 18 & 17 & 16 & 15 & 14 & 13 \\ \hline \text{面积} & 24 & 46 & 66 & 84 & 100 & 114 & 126 & 136 & 144 & 150 & 154 & 156 \\ \hline \end{array}$ 最大的可能面积是 $13\times 12=156$，最小的是 $1\times 24=24$，因此差值为 $156-24=132$ 平方单位。注意：当两个数的和固定时，它们的乘积会随着这两个数彼此越接近而增大。这意味着在周长相同的情况下，正方形的面积比任何长方形都大；而形状越接近正方形的长方形，其面积也会比其他周长相同的长方形更大。

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