Respuesta :
Answer:
The dimensions will minimize the amount of paper used is with length 9 inch and width 18 inch.
Step-by-step explanation:
Given : Johnny is designing a rectangular poster to contain 18 in² of printing with a 3-in margin at the top and bottom and a 6-in margin at each side.
To find : What overall dimensions will minimize the amount of paper used?
Solution :
Let l be the length and w be the width.
Area is [tex]A=l\times w[/tex]
Johnny is designing a rectangular poster to contain 18 in² of printing
i.e. Area of the inner printed section [tex]18=lw[/tex] ....(1)
A 3-in margin at the top and bottom and a 6-in margin at each side.
i.e. l'=l+6 and w'=w+12
Total Area of the Poster is [tex]A=(l+6)(w+12)[/tex] ....(2)
Substitute the value of l from (1) in (2),
[tex]A=(\frac{18}{w}+6)(w+12)[/tex]
[tex]A=18+\frac{216}{w}+6w+72[/tex]
[tex]A=90+\frac{216}{w}+6w[/tex]
[tex]A=90+\frac{216}{w}+6w[/tex]
Differentiate w.r.t w,
[tex]A'=-\frac{216}{w^2}+6[/tex]
For critical value put A'=0,
[tex]-\frac{216}{w^2}+6=0[/tex]
[tex]\frac{216}{w^2}=6[/tex]
[tex]w^2=\frac{216}{6}[/tex]
[tex]w^2=36[/tex]
[tex]w=6[/tex]
Differentiate again w.r.t w,
[tex]A''(w)=2(\frac{216}{w^3})[/tex]
[tex]A''(6)=2(\frac{216}{6^3})=2>0[/tex]
It is minimum at w=6.
Substitute in (1),
[tex]18=6l[/tex]
[tex]l=3[/tex]
The dimensions required are
l'=l+6= 3+6=9 inch
w'=w+12= 6+12=18 inch
The dimensions will minimize the amount of paper used is with length 9 inch and width 18 inch.
