Question

Optimization Problem:

Ronnie is designing a poster to contain 50in^2 of printing with margins of 4 inches each at the top and at the bottom and margins of 2 inches at each side. What overall dimensions will minimize the amount of paper used?

Answer 1

Let printing width = w and let printing length = x
x * w = 50
w = 50/x
Overall area A = (w + 4)(x + 8)
Sustituting for w gives:
A = (50/x + 4)(x + 8)
Multiplying out gives:
A = 4x + 400/x + 82
We need to find the value of x that makes A a minimum.

Are you following so far?

Answer 2

yeah, thanks

Answer 3

wait, are you sure you got the multiplication right for A? i got something else

Answer 4

just kidding, i did somethign wrong here

Answer 5

Gr8. To find the value of x to give the minimum value of A we differentiate A with respect to x:
dA/dx = 4 - 400/x^2
Now put the result equal to zero and solve for x:
4 - 400/x^2 = 0
x^2 = -400/-4 = 100
\[x=\sqrt{100}=10\]

Answer 6

that is the correct answer, thank you sir

Answer 7

my problem is i can't figure out the relevant equation to derive etc.

Answer 8

So the optimum printing length is 10 inches.
The optimum printing width is found from:
x * w = 50
10 * w = 50
w = 5
So the optimum overall dimensions will be:
Length = (10 + 8) = 18 inches
Width = (5 + 4) = 9 inches

Answer 9

could you do another one for me? or just part of it, i have it figured out but the arithmetic was crazy, i had to use wolfram, maybe there was another simpler way?

Answer 10

Do you follow the method to get the overall dimensions?

Answer 11

actually it's just 5" x 10"

Answer 12

we included the area of the print in the overall dimensions

Answer 13

A = (w + 4)((50/w) + 8)
that's print area + margins = overall area

Answer 14

Sorry for error. You are right. Good work :)