Answer:
[tex]A_p_r_i_n_t_e_d(w)=5-0.14w[/tex]
Explanation:
Take a look at the picture I attached you. The Rectangle area is given by:
[tex]A=w*h[/tex]
Where:
[tex]w=width\hspace{3}of\hspace{3}the\hspace{3}rectangle\\h=height\hspace{3}of\hspace{3}the\hspace{3}rectangle[/tex]
The total area of the poster is 5m^2, so:
[tex]5=w*h[/tex]
Isolating h:
[tex]h=\frac{5}{w}[/tex] (1)
Now, the printed area would be the product between w and h minus the length of the margins, therefore:
[tex]A_p_r_i_n_t_e_d=w*(h-2*0.07)[/tex] (2)
Replacing (1) in (2) and applying distributive property:
[tex]A_p_r_i_n_t_e_d(w)=w*(\frac{5}{w} -0.14)=5-0.14w[/tex]