Sameer Chotrani
Last Activity: 7 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
