# A poster is to contain 72 cm2 of printed matter with border of 4cm each at the top and bottom and 2cm on each side .find the dimensions if the total area of the poster is minimum

Sameer Chotrani
24 Points
5 years ago
Let the length and width of the poster excluding the margins be x and y respectively.

According to the question,
xy=72, i.e., y=72/x …............(1)

Total length of poster (l) = 4+x+4 = (x+8) cm
Total width of poster (b) = 2+y+2 = (y+4) cm

Total area of poster (A) = lb
i.e., A = (x+8)(y+4)
i.e., A(x) = (x + 8)(72/x + 4)                 [from (1)]
i.e., A(x) = 72 + 4x + 592/x +32
i.e., A(x) = 4x + 576/x + 104
i.e., A’(x) = 4 – 576/x2 + 0
i.e., A’(x) = (4x2 – 576)/x2 ….................(2)

At critical points,
A’(x) = 0
i.e., (4x2 – 576) = 0            [from (2)]
i.e, x2 = 144
i.e., x = 12 [ignoring negative square root since edge length can’t be nagative]
Putting in (1):
y = 72/12 = 6

To prove that this is a point of local minima you can use the 2nd derivative test as follows:
A’’(x) = 1152/x3          [from using the equation above (2)]
i.e., A’’(12) = 1152/(12)3 > 0
This shows that local minima exists at x= 12.

Therefore, required dimensions are
Length (l) = x+8 = 12+8 = 20cm
Breadth (b) = y+4 = 6+4 = 10cm