Question

Question 10 A closed rectangular box with a volume of 108ft is made from two kinds of materials. The top and bottom are made of material costing 56 cents per square foot and the sides from material costing 14 cents per square foot. Use Lagrange multipliers to find the dimensions of the box so that the cost of materials is minimized. Assume that w<l. 1 = ft W = ft h= ft

Answer 1

let the top and bottom of the box have dimensions a & J ft let the height box be H. of the I the box Then the cost cost of material for making will be Clas,z)= 227 (0.56) + (227+ 2yz) (0.14) =cCay,z) - 121207 + 0.28 (az+yz) C(2, Z) = 1.12xy +0.282 (x+y) - © We want to find the values of minimize the of the above function st. c. the volume XI, Z to cost és 108 fts ie v=xyz = 108. 2 = 1, let g(x,], z) = xyz - 108 = 0 so Cat 1. 12y + 0.28Z Cy = 1.122 +0.28Z Cz - 0.28 (0+3) & ga y z q= az gz 39

tuse fellow: ble Lagrange's Multiplier as x = dgx by = dgy ( = 192 =)1.12y +0.287 = dy I 1. 12x +0.28z = dorz & 0.28% +0.284 = doje & 1.12 2y +0.2871 - deyd 1. 12 ay t 0.28 y z dxyz & 0:28 22 +0.28 y1 = AXJL & From the of above equations 1.1264 + 0.2831 = '1.1987 +0.28yE first two 1 5 Loc => above From the first and third q equations 1. 1284 +0.2862=0.2361 +0.28 yz = 1.12 xq = 0.284 Z =) x = I 0.28 1.12 I Z и = x = OR L=hx So, use = 108 now

=) xyz - 108 = (x)(a) (4x) = 108 =) 4x² = 40x² = 108 3 x= 108 4 3=27 x=3 =) y=3 & 2 = 4x ch(3) - 12 so 3 ft W = 3 ft h = 12 ft to Hope this is helppull og