Question

The top and bottom margins of a poster of each 6 cm and the side margins are each 4 cm. If the area of the printed material on the poster is 384 sq.cm. Find the dimensions of the poster with the smallest area.

Answer 1

Hint: let \[x\] be the width of the printed area and \[y\] be the height of the printed area. Fixed area is given. Using the given condition we are going to find one of the variables and then the other. Here dimensions means height and width of a printed area. This problem comes under the application of derivatives.

Complete step-by-step answer:
Let \[x\] and \[y\] be the width and height of the printed area respectively.
Area is multiplication of width and height \[ \Rightarrow xy = 384\;c{m^2}\]
Then, \[ \Rightarrow y = \dfrac{{384}}{x}\;c{m^2}\] . ---- (a)
The total height of the poster including 6 cm margins at the top and bottom is \[y + 12\] .
The total width of the poster including 4 cm margins at sides is \[x + 8\] .
Hence, the area is, \[A = (x + 8)(y + 12)\]
We know the value of \[y\] , substituting we get,
 \[ \Rightarrow A = (x + 8)\left( {\dfrac{{384}}{x} + 12} \right)\]
 \[ \Rightarrow A = x\left( {\dfrac{{384}}{x} + 12} \right) + 8\left( {\dfrac{{384}}{x} + 12} \right)\]
 \[ \Rightarrow A = 12x + \dfrac{{3072}}{x} + 480\]
We need to find \[A'\] , i.e., differentiate with respect to \[x\] and equate it to zero ( \[A' = 0\] ).
We differentiate the function to minimize it.
 \[ \Rightarrow A' = 12 - \dfrac{{3072}}{{{x^2}}}\] (Since differentiation of constant term is zero)
 \[ \Rightarrow 0 = 12 - \dfrac{{3072}}{{{x^2}}}\]
 \[ \Rightarrow 12 = \dfrac{{3072}}{{{x^2}}}\]
 \[ \Rightarrow {x^2} = \dfrac{{3072}}{{12}}\]
 \[ \Rightarrow {x^2} = 256\]
 \[ \Rightarrow x = 16\]
Where \[x\] is width, it has to be positive.
 \[x = 16\;cm\]
To find \[y\] , substitute \[x = 16\;cm\] in (a) we get,
 \[ \Rightarrow y = \dfrac{{384}}{{16}} = 24\;cm\]
The total poster width is \[ \Rightarrow x + 8 = 16 + 8 = 24\;cm\]
The total porter height is \[ \Rightarrow y + 12 = 24 + 12 = 36\;cm\]
Hence, dimensions of the poster with the smallest area is \[24\;cm\] and \[36\;cm\] .
So, the correct answer is “ \[24\;cm\] and \[36\;cm\]”.

Note: In the above calculation part we did only simple multiplication and substitution. We solved this by optimizing the equation using differentiation and so that the area of the poster will be minimized. Always check your values are in the same unit. If not convert it. The procedure is the same for all the problems with different values.