1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Let's try the actual solution of the system
<span>y = –2x + 1
y = –2x – 3
Note that the slopes of the graphs of these two lines are the same: -2.
That means that the lines are parallel to one another.
Only the y-intercepts (1 and -3) are different.
Since the 2 lines never intersect, the system has no solution.
</span>
Change the order of integration.
Substitute 
and 
.
Answer:
D: x = 2
Step-by-step explanation:
Go down the multiple choices answers and plug each value in for x until the inequality is true:
A:
1/4(32) < 8
1/4 × 32/1 < 8
32/4 < 8
8 < 8
8 is not less than 8 so this is false (8 = 8).
B:
1/4(35) < 8
35/4 < 8
8 3/4 < 8
8 3/4 is not less than 8 so this is false (8 3/4 > 8).
C:
1/4(34) < 8
34/4 < 8
8 2/4 < 8
8 2/4 is not less than 8 so this is false (8 2/4 > 8).
D:
1/4(2) < 8
2/4 < 8
2/4 is less than 8 so this is true (2/4 = 1/2 = 0.5 > 8).
