# A sheet of paper is to contain \[18c{m^2}\] 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 which require the least amount of paper.

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**Hint:**Here we will firstly assume the dimension of the printed area on the paper. Then we will write the dimension of the paper with the addition of the margins. Then we will find the area of the paper and differentiate it and equate it to zero to get the value of the dimension of the paper.

**Complete step-by-step answer:**Let \[x\] be the length of the paper and \[y\] be the breadth of the paper in which the matter is printed.

It is given that the margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Therefore the dimension of the actual paper will be \[x + 4\] and \[y + 2\].

It is given that the sheet of paper is to contain \[18c{m^2}\] of printed matter. Therefore, we get

\[ \Rightarrow x \cdot y = 18\]………………………..\[\left( 1 \right)\]

Now we will find the area of the paper. Therefore, we get

\[A = \left( {x + 4} \right) \cdot \left( {y + 2} \right) = xy + 2x + 4y + 8\]

By using the equation \[\left( 1 \right)\] in the above equation we will write the area of the paper in terms of \[x\] only. Therefore, we get

\[ \Rightarrow A = 18 + 2x + 4\left( {\dfrac{{18}}{x}} \right) + 8\]

Simplifying the expression, we get

\[ \Rightarrow A = 2x + \dfrac{{72}}{x} + 26\]

To find the dimensions of the sheet which require the least amount of paper or to maximize the printed area we will differentiate the equation of the area of the paper and equate it to zero. Therefore, we get

\[2 + \dfrac{{ - 72}}{{{x^2}}} + 0 = 0\]

Now we will solve this equation to get the value of the \[x\].

Simplifying the equation, we get

\[ \Rightarrow \dfrac{{72}}{{{x^2}}} = 2\]

On cross multiplication, we get

\[ \Rightarrow {x^2} = \dfrac{{72}}{2} = 36\]

Taking square root on both sides, we get

\[ \Rightarrow x = \sqrt {36} = 6cm\]

Now by putting the value of \[x\] in the equation \[\left( 1 \right)\] to get the value of \[y\], we get

\[6 \cdot y = 18\]

\[ \Rightarrow y = \dfrac{{18}}{6} = 3cm\]

**Hence, the sheet which requires the least amount of paper is \[x + 4\] and \[y + 2\] which is \[10cm\] and \[5cm\].**

**Note:**We have to remember that the area of a rectangle is equal to the product of the length of the rectangle and the breadth of the rectangle. The area is measured in square units. To find the maximum area of utilization we have to differentiate the equation and equate it to zero to get the value of the variable corresponding to the maximum area of utilization. We should always write the units of the measurement of the variables.

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