An aquarium is to have a volume of 96 m^3. The length of its base is twice the width. Material for the base costs $12 per m^2; material for the sides costs $5 per m^2. Find the cost of the material for the cheapest such aquarium.
2006-11-13 15:05:04 · 3 answers · asked by JEFFREY S 1 in Science & Mathematics ➔ Mathematics
V = 2hw², h = 48/w²
Ab = 2w²
As = 2wh
Ae = wh
Al = 2(2wh + wh) = 6wh = 6*48/w
C = 12*2w² + 5*6*48/w
dC/dw = 48w - 1,440/w² = 0
48w = 1,440/w²
w³ = 30
w = 3.1072 m
h = 48/3.1072²
h = 4.9716 m
C = $781.44
2006-11-13 15:40:17 · answer #1 · answered by Helmut 7 · 1⤊ 1⤋
Make a graph with width as the x axis Plot costs on the y axis. Two lines: base and sides. Hmm, needs more thought....
2006-11-13 23:19:31 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
Minimize C = w * 2w * $12 + 2 * (w + 2w) * h * $5 subject to w * 2w * h = 96.
2006-11-13 23:15:17 · answer #3 · answered by Charles G 4 · 0⤊ 0⤋