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Home Partial Differentiation Maxima and Minima Examples Example 7
 
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Example 7

Find the dimensions of the box of volume one without a top which has the smallest area (if there is one). The box is sketched in Figure 11.7.13.

11_partial_differentiation-482.gif

Figure 11.7.13

Step 1

Let x, y, and z be the dimensions of the box, with z the height. We want the minimum of the area

A = xy + 2xz + 2yz given that xyz = 1.

Eliminating z, we have z = 1/(xy)

11_partial_differentiation-480.gif

The domain is the open region x > 0, y > 0 (see Figure 11.7.14).

 

11_partial_differentiation-483.gif

Figure 11.7.14

Step 2

11_partial_differentiation-481.gif

Step 3

11_partial_differentiation-486.gif

The critical point is11_partial_differentiation-487.gif where A = 22/3 + 2·2-1/3 + 2·2-1/3 = 22/3 + 25/3.

Step 4

Take for E the hyperreal region ε ≤ x ≤ H, ε ≤ y ≤ H where ε is positive infinitesimal and H is positive infinite.

Step 5

Let (x, y) be a boundary point of E. As we can see from Figure 11.7.15, there are four possible cases.

Case 1

x is infinitesimal. Then A is infinite because 2/x is.

Case 2

y is infinitesimal. A is infinite because 2/y is.

Case 3

x is not infinitesimal and y is infinite. A is infinite because xy is.

Case 4

y is not infinitesimal and x is infinite. A is infinite because xy is.

11_partial_differentiation-488.gif

Figure 11.7.15

CONCLUSION

A is infinite and hence greater than 22/3 + 25/3 on the boundary of E. Therefore A has a minimum at the critical point

11_partial_differentiation-484.gif

The box has dimensions

11_partial_differentiation-485.gif
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Last Update: 2006-11-15