Let's use the variable x to represent the length of the paper and y to represent the width of the paper.
If there are margins of 1 inch along the sides and 3 inches along the top and bottom, and the area with printer matter is 27 in², we can write the equation:
[tex](x-2)\cdot(y-6)=27[/tex]
Solving for x, we have:
[tex]\begin{gathered} x-2=\frac{27}{y-6} \\ x=\frac{27}{y-6}+2 \end{gathered}[/tex]
Then, the equation that we want to minimize is the area equation, so:
[tex]\begin{gathered} A=x\cdot y \\ A=(\frac{27}{y-6}+2)\cdot y \\ A=(\frac{27+2(y-6)_{}}{y-6})y_{} \\ A=(\frac{27+2y-12}{y-6})y \\ A=(\frac{2y+15}{y-6})y \\ A=\frac{2y^2+15y}{y-6} \end{gathered}[/tex]
The critical points of this function are:
y = 0 (one of the roots)
y = -15/2 (one of the roots)
y = 6 (function is not defined)
