Answer:
[tex]80\ \text{cm}\times 48\ \text{cm}[/tex]
Step-by-step explanation:
A = Area of the poster = [tex]6000\ \text{cm}^2[/tex]
x = Length of poster
y = Width of poster
[tex]xy=6000\\\Rightarrow y=\dfrac{6000}{x}[/tex]
Printed area would be
[tex](x-20)(y-12)\\ =(x-20)(\dfrac{6000}{x}-12)\\ =6000-12x-\dfrac{120000}{x}+240[/tex]
So we get the printable area as function
[tex]f(x)=6000-12x-\dfrac{120000}{x}+240[/tex]
Differentiating with respect to x we get
[tex]f'(x)=-12+\dfrac{120000}{x^2}[/tex]
Equating with zero
[tex]0=-12+\dfrac{120000}{x^2}\\\Rightarrow 12=\dfrac{120000}{x^2}\\\Rightarrow x^2=\dfrac{120000}{12}\\\Rightarrow x^2=10000\\\Rightarrow x=100[/tex]
Printed area length would be [tex]100-20=80\ \text{cm}[/tex] and breadth would be [tex]\dfrac{6000}{100}-12=60-12=48\ \text{cm}[/tex]
The dimensions that maximize the printed area has a length of 120 cm and width of 72 cm
Let x represent the length of the poster and y represent the width of the poster, hence:
xy = 6000
y = 6000/x
Since margins of 10 cm wide on the top and bottom and 6 cm wide on the sides hence:
length = x + 10 + 10 = x + 20, width = y + 6 + 6 = y + 12
The Area (A) = (x + 20)(y + 12)
A = xy + 12x + 20y + 240
A =6000 + 12x + 20(6000/x) + 240
A = 6240 + 12x + 120000/x
The maximum area is at dA/dx = 0:
dA/dx = 12 - 120000/x²
120000/x² = 12
12x² = 120000
x² = 10000
x = 100 cm
y = 6000/x = 6000/100 = 60 cm
length = 100 + 20 = 120 cm, width = y + 12 = 60 + 12 = 72 cm
The dimensions that maximize the printed area has a length of 120 cm and width of 72 cm
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