Answer:
The total height and width of the poster is approximately 28.93 and 38.57.
Step-by-step explanation:
Consider the provided information.
Let us consider the width and length of the printed part of the poster is x and y respectively.
For better understanding refer the figure 1:
The area of the printed part is 382 cm²
Area = xy = 382
y = 382/x
Now the total height of the poster is y + 6 + 6 = y + 12
Total width of the poster is x + 8 + 8 = x + 16
Thus the total area of the poster is:
A = (x + 16)(y + 12)
Now substitute the value of y in above equation.
[tex]A = (x+16)(\frac{382}{x}+12)[/tex]
[tex]A = 382+12x+\frac{6112}{x}+192[/tex]
[tex]A = 12x+\frac{6112}{x}+574[/tex]
Now differentiate the above equation with respect to x.
[tex]A' = 12-\frac{6112}{x^2}[/tex]
Now, substitute A'= 0 and solve for x.
[tex]0 = 12-\frac{6112}{x^2}[/tex]
[tex]12=\frac{6112}{x^2}[/tex]
[tex]x^2=\frac{6112}{12}[/tex]
[tex]x=22.57[/tex] Ignore the negative value of x as width can't be a negative number.
Now find A''
[tex]A'' = \frac{2(6112)}{x^3}[/tex]
Here A" is positive for x>0 and x = 22.57 is minimum.
Use the value of x to find all the respective dimensions of the poster.
Substitute the value of x in y = 382/x
y = 382/22.57 = 16.93
The total height of the poster is y + 6 + 6 = y + 12 = 16.93 + 12 = 28.93
Total width of the poster is x + 8 + 8 = x + 16 = 22.57 + 16 = 38.57
Hence, the total height and width of the poster is approximately 28.93 and 38.57.
Answer:
38.57 cm × 28.93 cm
Step-by-step explanation:
Let x be the length of the poster ( in cm ) and y be the height ( in cm ) of the poster,
Then the area of the poster,
A = x × y
∵ The top and bottom margins of a poster are 6 cm and the side margins are each 8 cm,
Thus, the total area of the poster other than printed material = 2(6x) + 2(8y) - 4(48) ( shown in the below diagram )
= 12x + 16y - 192
We have the printed area = 382 cm²
So, the area of the poster = printed area + non printed area
= 382 + 12x + 16y - 192
= 190 + 12x + 16y
∵ Area of the poster, A = xy
⇒ xy = 190 + 12x + 16y
⇒ y(x-16) = 190 + 12x
⇒ [tex]y=\frac{190+12x}{x-16}----(1)[/tex]
Area of the poster,
[tex]A(x)=x(\frac{190+12x}{x-16})[/tex]
[tex]=\frac{190x+12x^2}{x-16}[/tex]
Differentiating with respect to x,
[tex]A'(x)=\frac{(x-16)(190+24x)-(190x+24x^2)(1)}{(x-16)^2}[/tex]
[tex]=\frac{12x^2-384x-3040}{(x-16)^2}[/tex]
Again differentiating with respect to x,
[tex]y''=\frac{12224}{(x-16)^3}[/tex]
For maxima or minima,
y'=0
[tex]\implies \frac{12x^2-384x-3040}{(x-16)^2}=0[/tex]
[tex]\implies 12x^2-384x-3040=0[/tex]
[tex]\implies x\approx 38.57\text{ or }x\approx -6.57[/tex]
Since, dimension can not be negative,
If x = 38.57,
A''(x) = positive
Hence, A(x) is minimum,
From equation (1),
[tex]y=\frac{190+12(38.57)}{38.57-16}=28.93[/tex]
Therefore, the required dimensions are 38.57 cm × 28.93 cm
