Iestion 1 of 9 A realtor wishes to enclose $1050 \mathrm{~m}^{2}$ of land in a rectangular plot and then divide it into two plots with a fence parallel to one of the sides as shown in the figure. Let $x$ denote length of the large side and $y$ denote short side of the plot. What are the dimensions of the rectangular plot that require the least amount of fencing? (Use symbolic notation and fractions where needed.)

Step 1 ：The total area of the rectangular plot is given as 1050 m². This can be represented as the product of the length and the width, or \(x*y = 1050\).

Step 2 ：The total length of the fencing is the perimeter of the rectangle plus the length of the fence dividing the plot into two, which is \(2x + 3y\).

Step 3 ：We want to minimize the amount of fencing, so we need to minimize the function \(2x + 3y\).

Step 4 ：We can express y in terms of x using the area equation: \(y = 1050/x\).

Step 5 ：Substitute \(y = 1050/x\) into the fencing function: \(2x + 3(1050/x) = 2x + 3150/x\).

Step 6 ：To find the minimum of this function, we can take the derivative and set it equal to zero.

Step 7 ：The derivative of \(2x + 3150/x\) is \(2 - 3150/x²\).

Step 8 ：Setting this equal to zero gives: \(2 - 3150/x² = 0\).

Step 9 ：Solving for x gives: \(x² = 3150/2 = 1575\).

Step 10 ：Taking the square root of both sides gives: \(x = \sqrt{1575} \approx 39.68 m\).

Step 11 ：Substitute \(x = 39.68\) into \(y = 1050/x\) to find y: \(y = 1050/39.68 \approx 26.45 m\).

Step 12 ：So, the dimensions that require the least amount of fencing are approximately 39.68 m by 26.45 m. This can be confirmed by checking that these dimensions give an area of 1050 m²: \(39.68 m * 26.45 m = 1050 m²\).

Step 13 ：The final answer is \(\boxed{39.68 m, 26.45 m}\).