The cabinet that will enclose the Acrosonic model D loudspeaker system will be rectangular and will have an internal volume of 2.4 ft3. For aesthetic reasons, the design team has decided that the height of the cabinet is to be 1.5 times its width. If the top, bottom, and sides of the cabinet are constructed of veneer costing 90¢/ft2 and the front (ignore the cutouts in the baffle) and rear are constructed of particle board costing 30¢/ft2, what are the dimensions of the enclosure that can be constructed at a minimum cost?
Accepted Solution
A:
Answer:Dimensions of cabinetx (wide) = 1.93 fty (hight) = 2.895 ftp (depth) =0.43 ftStep-by-step explanation:Dimensions of cabinety height x wide p dephFrom problem statementy = 1.5 x V = y * x * p V = 1.5*x²p but p = V/y*x p = 2.4/1.5 x² p = 1.6 / x²Then Area of top and bottom A₁ = 2*x*p ⇒ 2*x*1.6/x² A₁ = 3.2 /xAnd cost in $ C₁ = 0,9 * 3.2 /x ⇒ C₁ = 2.88/xArea of sides (front and rear not included)A₂ = 2*y *p A₂ = 3*x*1.6/x² A₂ = 4.8/xAnd cost in $ C₂ = 0.9 * 4.8 /x C₂ = 4.32 /xArea of front and rear A₃ =2* y*x A₃ = 2*1.5 *x² A₃ = 3x²And cost C₃ = 0.3 * 3/x² = 0.9/x²Total cost C(x) = C₁ + C₂ + C₃ C(x) = 2.88/x + 4.32/x + 0.9x²Taking derivativesC´(x) = -2.88/x² - 4.32 /x² + 0.9 xC´(x) = 0 -2.88/x² - 4.32/x² + 0.9 x = 0 -2.88 - 4.32 + 0.9 x³ = 0-7.2 + x³ = 0 x³ = 7.2 x = 1.93 ft y = 1.5*1.93 = 2.895 ft and p = 0.43 ft