A sheet of paper is to contain 18 cm^2 of printed matter. The margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Find the dimensions of the sheet that require the least amount of paper.

To solve this problem, we need to minimize the area of the sheet of paper while ensuring that the printed matter area is fixed at 18 cm². Let's break down the problem step by step.

### Step 1: Define Variables
Let the dimensions of the sheet of paper be:
- Length of the sheet = $L$ (in cm)
- Width of the sheet = $W$ (in cm)

Given that the margins are 2 cm at the top and bottom, and 1 cm on each side, the printed area will be:
- Printed length = $L-4$ (since 2 cm margin at the top and bottom)
- Printed width = $W-2$ (since 1 cm margin on each side)

The printed area is given as 18 cm², so:
$$(L-4)(W-2)=18$$
This is the first equation we will use.

### Step 2: Area of the Sheet
The total area of the sheet is simply:
$$\text{Area of the sheet}=L\times W$$
We aim to minimize this area.

### Step 3: Express the Width in Terms of Length
From the equation $(L-4)(W-2)=18$, we can solve for $W$ in terms of $L$. Expanding the equation:
$$L\times W-2L-4W+8=18$$
Simplifying:
$$L\times W-2L-4W=10$$
Now, solve for $W$:
$$L\times W-4W=2L+10$$
Factor out $W$:
$$W(L-4)=2L+10$$
Solving for $W$:
$$W=\frac{2L+10}{L-4}$$

### Step 4: Minimize the Area
Now that we have $W$ in terms of $L$, we can express the total area $A$ of the sheet as:
$$A=L\times \frac{2L+10}{L-4}$$
Simplify the expression:
$$A=\frac{L(2L+10)}{L-4}$$
To minimize the area, we will take the derivative of $A$ with respect to $L$ and set it equal to 0.

### Step 5: Take the Derivative of the Area Function
First, expand the numerator:
$$A=\frac{2L^2+10L}{L-4}$$
Now, apply the quotient rule to differentiate:
$$\frac{dA}{dL}=\frac{(L-4)(4L+10)-(2L^2+10L)(1)}{(L-4{)}^2}$$
Simplify the numerator:
$$(L-4)(4L+10)=4L^2+10L-16L-40=4L^2-6L-40$$
Now subtract $(2L^2+10L)$:
$$(4L^2-6L-40)-(2L^2+10L)=2L^2-16L-40$$
Thus:
$$\frac{dA}{dL}=\frac{2L^2-16L-40}{(L-4{)}^2}$$
Set the derivative equal to 0 to find the critical points:
$$2L^2-16L-40=0$$
Divide through by 2:
$$L^2-8L-20=0$$
Solve this quadratic equation using the quadratic formula:
$$L=\frac{-(-8)\pm \sqrt{(-8{)}^2-4(1)(-20)}}{2(1)}$$
$$L=\frac{8\pm \sqrt{64+80}}{2}=\frac{8\pm \sqrt{144}}{2}=\frac{8\pm 12}{2}$$
Thus, $L=\frac{8+12}{2}=10$ or $L=\frac{8-12}{2}=-2$.

Since length cannot be negative, we take $L=10$ cm.

### Step 6: Find the Corresponding Width
Substitute $L=10$ into the equation for $W$:
$$W=\frac{2(10)+10}{10-4}=\frac{20+10}{6}=\frac{30}{6}=5\text{cm}$$

### Step 7: Verify the Printed Area
The printed area is:
$$(L-4)(W-2)=(10-4)(5-2)=6\times 3=18{\text{cm}}^2$$
Thus, the printed area is correct.

### Step 8: Conclusion
The dimensions of the sheet that require the least amount of paper are:
- Length = 10 cm
- Width = 5 cm