First markdown: 0.75($500) = $375
Second markdown: 0.75($375) = $281.25
Third markdown: 0.75($281.25) = $210.94
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).
A: (x + 5i)^2
= (x + 5i)(x + 5i)
= (x)(x) + (x)(5i) + (5i)(x) + (5i)(5i)
= x^2 + 5ix + 5ix + 25i^2
= 25i^2 + 10ix + x^2
B: (x - 5i)^2
= (x + - 5i)(x + - 5i)
= (x)(x) + (x)(- 5i) + (- 5i)(x) + (- 5i)(- 5i)
= x^2 - 5ix - 5ix + 25i^2
= 25i^2 - 10ix + x^2
C: (x - 5i)(x + 5i)
= (x + - 5i)(x + 5i)
= (x)(x) + (x)(5i) + (- 5i)(x) + (- 5i)(5i)
= x^2 + 5ix - 5ix - 25i^2
= 25i^2 + x^2
D: (x + 10i)(x - 15i)
= (x + 10i)(x + - 15i)
= (x)(x) + (x)(- 15i) + (10i)(x) + (10i)(- 15i)
= x^2 - 15ix + 10ix - 150i^2
= - 150i^2 + 5ix + x^2
Hope that helps!!!
