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).
well, putting the movie in fifths, she watched 3/5, the whole movie will be 5/5 = 1.
![\begin{array}{ccll} \stackrel{movie}{fraction}&hour\\ \cline{1-2} \frac{3}{5}&\frac{5}{8}\\[1em] \underset{whole}{1}&x \end{array}\implies \cfrac{~~ \frac{3}{5}~~}{1}=\cfrac{~~ \frac{5}{8}~~}{x}\implies \cfrac{3}{5}x=\cfrac{5}{8}\implies \cfrac{3x}{5}=\cfrac{5}{8} \\\\\\ 24x=25\implies x = \cfrac{25}{24}\implies x = 1\frac{1}{24}\qquad \textit{1 hour, 2 minutes and 30 seconds}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccll%7D%20%5Cstackrel%7Bmovie%7D%7Bfraction%7D%26hour%5C%5C%20%5Ccline%7B1-2%7D%20%5Cfrac%7B3%7D%7B5%7D%26%5Cfrac%7B5%7D%7B8%7D%5C%5C%5B1em%5D%20%5Cunderset%7Bwhole%7D%7B1%7D%26x%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B~~%20%5Cfrac%7B3%7D%7B5%7D~~%7D%7B1%7D%3D%5Ccfrac%7B~~%20%5Cfrac%7B5%7D%7B8%7D~~%7D%7Bx%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B5%7Dx%3D%5Ccfrac%7B5%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B3x%7D%7B5%7D%3D%5Ccfrac%7B5%7D%7B8%7D%20%5C%5C%5C%5C%5C%5C%2024x%3D25%5Cimplies%20x%20%3D%20%5Ccfrac%7B25%7D%7B24%7D%5Cimplies%20x%20%3D%201%5Cfrac%7B1%7D%7B24%7D%5Cqquad%20%5Ctextit%7B1%20hour%2C%202%20minutes%20and%2030%20seconds%7D)
Answer:
sec theta = (sqrt24/5) cos theta = -2/5 tan theta = (-[sqrt 21]/2) sec theta = 5/2 csc theta = (5sqrt21)/21 cot theta = (-2sqrt21)/21
Step-by-step explanation:
During the problem, secx = -5/2, we can assume that as cos = -2/5. -2 = x. 5 = r. find for Y with: x^2+y^2=r^2. After that, plug in for the variables and you get all the answers. Rationalize the square roots, don't forget.
Answer:
Steps and solution in the attached picture.
Step-by-step explanation:
Steps and solution in the attached picture.
Answer:
x=-2
Step-by-step explanation:
add 5x , then subtract 45, then divide by 15
