\"A\" is the exponent symbol A rectangular container with an open top is required to have a volume of 24 cubic meters. Also, one side of the rectangular base is required to be 4 meters long. If the material for the base costs $3 per square meter, and the material for the side costs $2 per square meter, nd the dimensions of the container so that the cost of material to make it will be a minimum. Steprl} Make a diagram ' h V 4....-) J. are (bJWhnt formulas will be used in this problem? 1i|."'olun11e of cube = LWH = length*width*height In our question one side (length) = 4m, so volume= 4wh is = 24m'\"3 {given}, 24 = 4wh divide each side by 4, so 6 = Ell. l= 4m, Mb; 24, : 6 ( C) Figure out what the constraint is The constraint is width*height = 6 because of the volume, formulated in b part because one side is given as 4m, so wh= 6. wh= 6, L=4m The volume is the constraint. It has to be 24m^3 which also limits the length value of the width and height when solving for them. (d) What do you want to maximize or minimize? C represents cost Minimizing cost of base area + side wall area cost= base area*8 + side area*2 = 32w + 4wh+ 16h Now we know wh= 6, C = cost = (32w) + 24 +96/w cost = 8(4w)+2[2hw+2*4h] C =32w+4hw+16h C =32w + 4(24/4w)w + 16(6/w), h= 24/4w plug in for h and simplify, "w" cancels C =32w + 24 + 96/w Simplification of cost function and substitution of given values to solve for "w" which allows us to find the critical points which will find the minimum. Step 2 Write down formulas Formula of volume LWH = 24, 4 = L Cost function = c = 32w + 24 +96/w Cost = 32w + 24 + 96/w Step:3 Substitute into formula you want to maximize or minimize Minimize cost function (C = 32w + 24 + 96/w), for minimizing cost, we took the derivative of C to solve for critical points to then find test points. Next step taking derivative of cost function, so C'= 32w^2 - 96, then C' = 0 0 =32w 2 - 96 96 = 32w 2, divide by 32, 3 = w^2, square root, +/-V3 = wStep:4 Find the Critical points C' =32w 2-96 C'=32(w*2-3)=0, Factor 32 out and set equal to 0, solve for w w 2=3, square root W= +/- V3 Number line- The number line shows us our test points to find where the graph is increasing/decreasing to find the local extrema. Step:5 Test the critical points Test points are +/- v4 from the number line: 0 = 32(-4)^2 - 96 = 32, positive answer means increasing 0 = 32(V4)^2 - 96 = -224, negative means decreasing The critical points provide the maximum and minimum. This is an optimization problem where we must minimize the cost. The critical points show the local minimum and maximum of the graph which is needed to know the minimum value. Step:6 Answer the question To minimize the cost of the material, the dimensions of the container must be... L =4m W = V3 = 1.732m H =6/w =6/