poster dimensions: L+4; W+16
LW=380
areatotal=(L+4)(W+16)=(L+4)(380/L+ 16)
dareatotal/dl=0=(380/L + 16) + (L+4)(-380/L^2)
solve for L
0= 380L+16*L^2- 380L-4*380
L^2=380/4=95
L= sqrt 95
at this point recheck my math work, I have a headache. If correct, solve for W, then solve for the poster dimensions.
The top and bottom margins of a poster are 2 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 380 square centimeters, find the dimensions of the poster with the smallest area.
Width =
Height =
2 answers
wh = 380
a = (w+16)(h+4)
= (w+16)(380/w + 4)
= 4w + 444 + 6080/w
da/dw = 4 - 6080/w^2
minimum area when da/dw = 0. So solve that and evaluate w and h. Then the poster dimensions are w+16 and h+4.
a = (w+16)(h+4)
= (w+16)(380/w + 4)
= 4w + 444 + 6080/w
da/dw = 4 - 6080/w^2
minimum area when da/dw = 0. So solve that and evaluate w and h. Then the poster dimensions are w+16 and h+4.
1 answer