Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .
Step-by-step explanation:
We have , A page should have perimeter of 42 inches. The printing area within the page would be determined by top and bottom margins of 1 inch from each side, and the left and right margins of 1.5 inches from each side. let's assume width of the page be x inches and its length be y inches So,
Perimeter = 42 inches
⇒ [tex]2(x+y) = 42\\x+y = 21\\y = 21-x[/tex]
width of printed area = x-3 & length of printed area = y-2:
area = [tex]length(width)[/tex]
[tex]area = (x-3)(y-2)\\area = (x-3)(21-x-2)\\area = (x-3)(19-x)\\area = -x^{2} + 22x -57[/tex]
Let's find [tex]\frac{d(area)}{dx}[/tex]:
[tex]\frac{d(area)}{dx}[/tex] = [tex]\frac{d(-x^{2}+22x-57)}{dx} = -2x +22[/tex] , for area to be maximum [tex]\frac{d(area)}{dx}[/tex]= 0
⇒ [tex]-2x+22 = 0\\2x =22\\x=11 inches[/tex]
And ,
[tex]y = 21-x\\y = 21-11\\y = 10 inches[/tex]
∴ Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .
So, the page should be 11 inches wide and 10 inches long
It is the total distance covered by the rectangle around its outside.
Let the width of the page be [tex]x[/tex] inches
and its length be [tex]y[/tex] inches
Then we can write as,
[tex]2(x+y) = 42\\x+y = 21\\y = 21-x[/tex]
And the width of printed area =[tex]x-3[/tex]
length of printed area =[tex]y-2[/tex]
Then the area will be,
[tex](x-3)(y-2)= (x-3)(21-x - 2)\\= (x-3)(19-x)\\= -x^2 + 22x - 57[/tex]
Now, differentiating with respect to [tex]x[/tex]
[tex]\frac{dA}{dx} =-2x + 22\\=0\\2x=22\\x = 11[/tex]
Then [tex]y = 21-11 = 10[/tex]
The page should be 11 inches wide and 10 inches long
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