A carpenter worked 10 weeks on a particular job, 51/2 days per week and 7 3/4 hours per day. How many hours did the carpenter wo
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Answer:
The dimensions of the rectangular poster is 15 in by 5 in.
Step-by-step explanation:
Given that, the area of the rectangular poster is 75 in².
Let the length of the rectangular poster be x and the width of the rectangular poster be y.
The area of the poster = xy in².
....(1)
1 in margin at each sides and 3 in margin at top and bottom.
Then the length of printing space is= (x-2.3) in
=(x-6) in
The width of printing space is = (y-2.1) in
=(y-2) in
The area of the printing space is A =(x-6)(y-2) in²
∴ A =(x-6)(y-2)
Putting the value of y
Differentiating with respect to x
Again differentiating with respect to x
To find the minimum area of printing space, we set A' = 0
Now putting x=±15 in A''
Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.
From (1) we get
Putting the value of x
=5 in
The dimensions of the rectangular poster is 15 in by 5 in.