A rectangular poster is to contain 108cm^2 of printed matter with margins of 6cm at the top and bottom and 2 cm on the sides. what's the least cost to make the poster if the printed material costs 5 cents/cm^2 and the margins are 1 cent/cm^2?
Answers
Let x be the width in cm of the printed material on the poster and y the height.
Since we are supposed to have 108 cm2
of printer material,
xy = 108.
With the margins, the total area of the poster will be
A = (x + 4)(y + 12),
and since we are paying 20 cents/cm2
for the poster, the cost is
C = .20(x + 4)(y + 12),
and this is what we want to minimize.
First solve the equation xy = 108 in terms of y: y = 108/x (or solve it in terms of x if you
like). Then substitute this into C:
C = .20(x + 4)
108
x
+ 12
= 31.20 + 2.40x +
86.40
x
.
We want to minimize C so we take a derivative:
C
0 = 2.40 −
86.40
x
2
.
Now we want to find the critical numbers of C so we set C
0
equal to 0. If C
0 = 0 then
86.30
x2 = 2.40, so
x = ±6.
We can rule out the solution x = −6, and since the only remaining critical number is
x = 6, we can be assured that that value minimizes C.
Finally, we have to give the minimum cost, which is the value of C when we plug in x = 6:
.20(6 + 4)
108
6
+ 12
= 60