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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A poster is to contain $50cm^2$ of matter with borders of 4 cm at top and bottom and of 2 cm on each side. Find the dimensions if the total area of the poster is minimum.

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1 Answer

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Toolbox:
  • For greatest area, maximise area.
  • Working rule for maxima minima.
  • Find the function to be maximised or minimised
  • Convert the function into single variable
  • Apply the conditions for maximum or minimum i.e., first derivative = 0 ( w.r.to the variable converted) and second derivative positive for minima and negative for maxima.
  • Answer the question.
Step 1
Let the dimensions be $ x $ cm length and $ y$ cm breadth.
Area of poster = $ A = xy$
Step 2
Given : (x-8)(y-4)=50$
or $ y = \large\frac{50}{x-8}+4$
$ \Rightarrow A = \bigg[ \large\frac{50}{x-8}+4 \bigg]^x$
Step 3
$ \large\frac{dA}{dx}=4-\large\frac{400}{(x-8)^2}=0$
Ans $ \Rightarrow x = 18, \: \: y = 9$
Step 4
$ \large\frac{d^2A}{dx^2}=$ positive $ \Rightarrow A$ is minimum.
answered Apr 12, 2013 by thanvigandhi_1
 

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