Probem15+Max+S.+Paul+F.



My scanner's not working so bear with me as i describe the diagram. We have a rectangular box with length l, width w, and height h. The objective equation would be V=l*w*h Now we will substitute values in for l and h to get an equation that relates volume only to the width of the box. One of the givens was that l = 2w, which we can plug in. If we set our volume equation = to 10 meters cubed we can solve for h in terms of w and plug that into our objective equation. 10=l*w*h 10=2w*w*h h=5/2w^2

Now that we have l and h in terms of w we can create our objective equation for the cost of the box. First,we have to decide whether we are minimizing or maximizing, we are going to minimize the cost. Some givens: cost per square meter of base = 10 $, cost per square meter of side would be 6$

C = l*w*10 + w*h*6*2 + l*h*2*6 + l*w*6 C= 20w^2 + 60/w + 120/w +12w^2 C= 32w^2 +180/w

Now we must take the first derivative to find the critical values. C'=64w - 180/w^2 0=64w - 180/w^2 64w=180/w^2

w^3 = 180/64 w= 1.411 meters

Plug that number back into our equation for cost and get C= 191.27$

The domain for w would be over ( 0, infinity) And the limit of C as x approaches 0 is infinity; the limit of C as x approaches infinity is also infinity...so we have an absolute min at x = 1.411

Looking at the graph of C, we no it is parabolic and concave up because the first and highest term is 32w² The function is never undefined so therefore the critical value found above is the only one. The only place where C' would be equal to zero would be at w=1.411 meters, the parabloa's vertex, which is C(w)'s absolute minimum.