2
$\begingroup$

this is a sample final question for a multivariable calculus course.

"A rectangular box has two opposing sides (left and right) made of gold, two (front and back) of silver, and two (top and bottom) of bronze. The cost (per unit area) of the gold is $4dollars/m^2$, the silver is $2dollars/m^2$ and the bronze is $1dollar/m^2$. The box has volume $8m^3$ and each of its edges are no more than 8m long. What are the minimum and maximum costs of such a box.

Attempt at a solution

Cost of the box = $4\times2(x\times y)+2\times2(x \times z)+2(y\times z)$

$x\times y\times z = 8$

By using lagranges multiplier, we can find critical point (1, 2 ,4), which would have a cost of $48 (minimum cost)

However, since this is the only critical point, I am not sure how to proceed to get the maximum cost.

I tried introducing another constraint x + y + z < 24, but am not sure how to apply it.

Intuitively, I can guess that the dimensions are (8, 8, $1\over 8$) to maximize the surface with the gold, and would have a cost of $518

Any help/suggestions to get the maximum cost?

Thanks :)

$\endgroup$
1
  • $\begingroup$ I think you've found it. If there is only one critical point, then checking the boundaries against it should cover it. Which is what you did, maxing out the edge length of x and y at 8m. Just my guess. $\endgroup$ – turkeyhundt Dec 10 '14 at 7:42
0
$\begingroup$

Your intuition proves to be correct, but you would need to give something a bit more formal in the solution to an exam problem. Your proposed constraint isn't strong enough since there is nothing in the inequality that forbids any one (or even two) of the edge measurements to exceed 8 meters.

The given constraint requires that the solution(s) for the maximum cost lie inside the cube $ \ [ 0 \ , \ 8 ] \ \times \ [ 0 \ , \ 8 ] \ \times\ [ 0 \ , \ 8 ] \ $ . This constraint is essential since the one for the volume of the metal box, $ \ xyz \ = \ 8 \ $ , alone would also any one or two of the edge measurements to "run to infinity". The "hyperbolic" surface is open, so the cost function, which is a linear combination of pairs of length variables, could be made arbitrarily large.

[pic to be inserted]

the open surface xyz = 8 with the location of the minimal cost indicated

$ \ \ $

With the "maximum lengths" constraints, the volume constraint surface intersects the planes $ \ x \ = \ 8 \ \ , \ \ y \ = \ 8 \ \ , \ \ z \ = \ 8 \ $ at three hyperbolic curves $ \ yz \ = \ 1 \ \ , \ \ xz \ = \ 1 \ \ , \ \ xy \ = \ 1 \ $ , respectively. This now requires the variables to lie in the intervals $ \ \frac{1}{8} \ \le \ x \ \le \ 8 \ $ and similarly for $ \ y \ $ and $ \ z \ $ in order to work on all three curves simultaneously.

[second pic]

the constraint planes added, with the intersection hyperbolas visible and the location of the maximal cost indicated

$ \ \ $

The Lagrange-multiplier method will only locate the one point where the cost function $ \ C \ = \ 8 \ xy \ + \ 4 \ xz \ + \ 2 \ yz \ $ can be tangent to $ \ xyz \ = \ 8 \ $ , which occurs for $ \ C \ = \ 48 \ $ , as you found. There are no other places where the "cost surface" will be tangent to the open volume constraint surface for any value of $ \ C \ $ .

Instead, we are left to check the curves on the three individual planes. We find on the plane $ \ z \ = \ 8 \ $ , for instance, that the cost function becomes $ \ C \ = \ 8 \ xy \ + \ 4 \ x \ \cdot \ 8 \ + \ 2 \ y \ \cdot \ 8 \ $ . Together with the hyperbolic equation $ \ xy \ = \ 1 \ $ , we have $ \ C \ = \ 8 \ \cdot 1 \ + \ 32 \ x \ + \ 16 \ \frac{1}{x} \ $ ; but this reduced function has a positive second derivative at its "critical point", so this is not a maximum value for the cost function. We obtain similar results on the other two planes, $ \ y \ = \ 8 \ $ and $ \ z \ = \ 8 \ $ .

This leaves us to check the three hyperbolic curves at each of their two endpoints:

z = 8 :
$$ \ C \ = \ 8 \ (1) \ + \ 32 \ \cdot \ \frac{1}{8} \ + \ 16 \ \cdot \ 8 \ \ \ , \ \ \ C \ = \ 8 \ (1) \ + \ 32 \ \cdot \ 8 \ + \ 16 \ \cdot \ \frac{1}{8} \ \ ;$$

y = 8 : $$ \ C \ = \ 8 \ \cdot \ \frac{1}{8} \ \cdot \ 8 \ \ + \ 4 \ ( 1 ) \ + \ 2 \ \cdot \ 8 \ \cdot \ 8 \ \ \ , \ \ \ C \ = \ 8 \ \cdot \ 8 \ \cdot \ 8 \ + \ 4 \ (1) \ + \ 2 \ \cdot \ 8 \ \cdot \ \frac{1}{8}\ \ ;$$

x = 8 : $$ \ C \ = \ 8 \ \cdot \ 8 \ \cdot \ \frac{1}{8} \ + \ 4 \ \cdot \ 8 \ \cdot \ 8 \ + \ 2 \ (1) \ \ \ , \ \ \ C \ = \ 8 \ \cdot \ 8 \ \cdot \ 8 \ + \ 4 \ \cdot \ 8 \ \cdot \ \frac{1}{8} \ + \ 2 \ (1) \ \ . $$

It is clear from this that in order to maximize the cost, we want the "product of two lengths equal to 1" factors to be multiplied by the lowest coefficients, $ \ 4 \ $ and $ \ 2 \ $ . Consequently, we need to have the product $ \ xy \ = \ 8 \ \cdot \ 8 \ $ , that is, as we recognized intuitively, the gold faces of the box should be squares with side $ \ 8 \ $ meters. This is either the fourth or the sixth of the equations above, giving a maximal cost of $ \ 518 \ $ dollars. (In fact, there are only three distinct cost values at the six indicated endpoints. In the diagram above, we see that this occurs because pairs of "cost hyperbolas" meets at just three locations.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.