To maximize the printed area, the dimensions of the poster should be 43.960 inches by 17.584 inches.
The goal is to maximize the printed area, which is equivalent to maximizing the product of the length and width. However, we have constraints with margins. Let x be the length and y be the width.
xy = 773 so that;
y = 773/x...(1)
Also;
Area, A = (x - 5)(y - 2)
A = xy - 2x - 5y + 10...(2)
putting 773/x for y in equation (2) gives;
A = x(773/x) - 2(773/x) - 5(773/x) + 10
A = 773 - 2x - 3865/x + 10
A = 783 - 2x - 3865/x
taking the derivative we have;
A' = -2 - 3865(-1/x²)
A' = -2 + 3865/x²
-2 + 3865/x² = 0
3865/x² = 2
2x² = 3865 {cross multiplication}
x² = 3865/2
x² = 1932.5
x = 43.960
substituting 43.960 into equation (1) gives
y = 17.584.
Thus, for the given margins, the dimensions of 43.960 inches by 17.584 inches will result in a poster with the maximum possible printed area of 773 square inches.