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Answer:
the dimensions of the poster with the smallest area is 36cm by 54cm
Step-by-step explanation:
✓Let us represent the WIDTH of the printed material on the poster as "x"
✓Let us represent the HEIGHT of the printed material on the poster as "y"
✓ The given AREA is given as 864 cm2
Then we have
864 cm2= xy ...................eqn(1)
We can make "y" subject of the formula.
y= 864/x .......................eqn(2)
✓The total height the big poster which includes the 9cm margin that is at the bottom as well as the top is
(y+18)
✓The total width of the poster which includes the 6cm margin that is at the bottom as well as the top is
(x+12)
✓Then AREA OF THE TOTAL poster
A= (y+18)(x+12) ...................eqn(3)
Substitute eqn (2) into eqn(3)
A= ( 18+ 864/x)(x+12)
We can now simplify by opening the bracket, as
A=18x +1080 +10368/x
A= 18x +10368/x +1080
Let us find the first derivative of A which is A'
A'= 18-(10368/x²)
If we set A' =0
Then
0= 18- (10368/x²)
18= (10368/x²)
x²= 10368/18
x²= 576
x=√576
x=24
The second derivatives will be A"= 2(10368)/x³ and this will be positive for x> 0, and here A is concave up and x=24 is can be regarded as a minimum
The value of "y" when x=24 can now be be calculated using eqn(2)
y= 864/x
y= 864/24
y=36cm
✓The total width of the poster= (x+12)
= 24+12=36cm
✓The total height big the poster= (y+18)=36+18=54cm
the dimensions of the poster with the smallest area is 36cm by 54cm
Answer:
The total width of the paper [tex]=36 cm.[/tex]
The total height of the paper [tex]=54cm[/tex]
Step-by-step explanation:
Given information:
Top margin of the paper = 9 [tex]cm\\[/tex]
Bottom margin of the paper = 6 [tex]cm\\[/tex]
Area of the printed material = [tex]864[/tex] [tex]cm^2[/tex]
Let, the width of the printed material = [tex]x[/tex]
And the height of the printed material = [tex]y[/tex]
So, Area [tex]x \times y=864[/tex] [tex]cm^2[/tex]
After including margins;
Width of the paper [tex]= (x+12)[/tex]
Height of the paper [tex]= (y+18)[/tex]
Area [tex](A) = (y+18) (x+12)[/tex]
[tex]A=18x+(10368/x)+1080\\[/tex]
Take first derivative:
[tex]A'= 18- (10368/x^2)[/tex]
When [tex]A'=0[/tex]
Then,
[tex]18-(10368/x^2)=0\\x^2=576\\x=24[/tex]
Now ,when we take second derivative and check if it is positive or not ,
We find that it is grater than zero so the obtained value can be consider as minimum and can be proceed for further solution.
Hence ,
[tex]x \times y=864\\y=864/24\\y=36\\[/tex]
Now ,
The total width of the paper
[tex]= 24+12\\=36 cm.[/tex]
And , total height of the paper
[tex]=36+18\\=54 cm.[/tex]
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