Ok the question is: A poster is to have an area of 180in with a 1" margin on the bottom and sides and a 2" margin at the top. What dimension will give the largest printed area?
I'm just wondering if anyone could check over my work cause I am not sure whether or not I am doing the question correctly.
Ok to start with I drew the poster, than drew the printed material part.
I let the printed material area be = L*W
I let the poster area be = (L+3)(W+2)
this gave me
180 = (L+3)(W+2)
I solved for W to obtain:
(180)/(L+3) - 2 = W
Subbing this into my printed area equation gave me:
A = (L)(180/(L+3) - 2)
A = (180L)/(L+3) - 2L
A' = [(180)(L+3) - (180L)(1)]/ (L+3)^2 - 2
A' = 540 / (L+3)^2 - 2
Solving for zero gave me:
540 / (L+3)^2 - 2 = 0
540 / 2 = (L+3)^2
L = sqrt[(270) - 3]
L = 3[sqrt]30 - 3 = 13.43
Than I sovled for W to get:
(180)/(13.43+3) - 2 = 8.95
so i get a dimension of 8.95 * 13.43
The reason I am curious is becasue every other question before this had "nice" numbers with no decimals, so this just seemed out of place.
I'm just wondering if anyone could check over my work cause I am not sure whether or not I am doing the question correctly.
Ok to start with I drew the poster, than drew the printed material part.
I let the printed material area be = L*W
I let the poster area be = (L+3)(W+2)
this gave me
180 = (L+3)(W+2)
I solved for W to obtain:
(180)/(L+3) - 2 = W
Subbing this into my printed area equation gave me:
A = (L)(180/(L+3) - 2)
A = (180L)/(L+3) - 2L
A' = [(180)(L+3) - (180L)(1)]/ (L+3)^2 - 2
A' = 540 / (L+3)^2 - 2
Solving for zero gave me:
540 / (L+3)^2 - 2 = 0
540 / 2 = (L+3)^2
L = sqrt[(270) - 3]
L = 3[sqrt]30 - 3 = 13.43
Than I sovled for W to get:
(180)/(13.43+3) - 2 = 8.95
so i get a dimension of 8.95 * 13.43
The reason I am curious is becasue every other question before this had "nice" numbers with no decimals, so this just seemed out of place.
