The volume of the box must be 100 cubic meters. the cost of the material to be used on the sides is $11 per square meter and the cost of the material to be used on the bottom is $19 per square meter. what is the minimum cost?

   V = Lwh 
                  10 = (2w)(w)(h) 
                  10 = 2hw^2 
                    h = 5/w^2

Cost:    C(w) = 10(Lw) + 2[6(hw)] + 2[6(hL)]) 
                       = 10(2w^2) + 2(6(hw)) + 2(6(h)(2w) 
                       = 20w^2 + 2[6w(5/w^2)] + 2[12w(5/w^2)] 
                       = 20w^2 + 60/w + 120/w 
                       = 20 w^2 + 180w^(-1) 
             C'(w) = 40w - 180w^(-2)

Critical numbers: 
           (40w^3 - 180)/w^2 = 0 
                       40w^3 -180 = 0 
                               40w^3 = 180 
                                   w^3 = 9/2 
                                       w = 1.65 m 
                                        L = 3.30 m 
                                        h = 1.84 m

Cost:  C = 10(Lw) + 2[6(hw)] + 2[6(hL)]) 
               = 10(3.30)(1.65) + 2[6(1.84)(1.65)] + 2[6(1.84)(3.30)]) 
               = $165.75                       cheapest cost

answer: c i think

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3