Answer:
a. contrapositive because it's the converse and inverse. True.
b. converse because it's the reverse of conditional statement. True.
c. That is false so it's not converse, inverse, or contrapositive.
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).
Number of weekend minutes used: x
Number of weekday minutes used: y
This month Nick was billed for 643 minutes:
(1) x+y=643
The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44
We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44
Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y
Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44
Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72
Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243
Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400
Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
Hi!
When you use postulates and theorems, you need to make sure to only use the given information that you know. Look for the given statements, and congruence marks on the figure. Those are also considered given.
By looking, you are given an angle and a side. The side comes first. SU=TV.
So, that makes it so Side-Angle-Side would be the best option.
I hope this helps!
