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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?

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

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Other answers:

just kidding, i did somethign wrong here
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 \[x=\sqrt{100}=10\]
that is the correct answer, thank you sir
my problem is i can't figure out the relevant equation to derive etc.
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
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?
Do you follow the method to get the overall dimensions?
actually it's just 5" x 10"
we included the area of the print in the overall dimensions
A = (w + 4)((50/w) + 8) that's print area + margins = overall area
Sorry for error. You are right. Good work :)

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