Answer:
25.920 W
Step-by-step explanation:
Fill in the given number and do the arithmetic.
P = f(12) = 0.015(12^3) = 25.920 . . . . . watts
Check out the attached image. I drew what I think your book is showing. The figure on the left is triangle ABC without any extended segments. The figure on the right has segment AB extended shown in red. This forms the exterior angle x
The rule that connects x, y and z together is the remote interior angle theorem. It says that adding two interior angles is going to be equal to the exterior angle that is not touching either interior angle. The "remote" part means "far away" so just think of the two angles that are furthest way or not touching the exterior angle in question.
In terms of algebra, the rule is
x+y = z
Given that the number is C.
The number divided by 7 is C/7
Difference means we will subtract 9 from C/7.
Therefore, the equation will be:
C/7 - 9 = 2 (multiply all terms by 7 to get rid of the fraction)
C - 63 = 14
C = 14 + 63 = 77
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).
