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**masters** Area of paper = $\displaystyle 11\left(\frac{17}{2}\right)=\frac{187}{2} \ \ in^2$

Area of the printed part = $\displaystyle A(x)=\left(\frac{17}{2}-2x\right)\left(11-2x\right)=\frac{187}{2}-39x+4x^2$

Area of the printed part cannot exceed area of the paper, so

$\displaystyle 4x^2-39x+\frac{187}{2}<\frac{187}{2}$

Domain of Area A = $\displaystyle \left\{0<x<\frac{39}{4}\right\}$

Range of Area A = $\displaystyle \left\{0<A(x)<\frac{187}{2}\right\}$

For part (c), just substitute these values in A(x).