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).
Answer:
0.5
Step-by-step explanation:
Solution:-
- The sample mean before treatment, μ1 = 46
- The sample mean after treatment, μ2 = 48
- The sample standard deviation σ = √16 = 4
- For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation.
Cohen's d = 
- Where, the pooled standard deviation (sd_pooled) is calculated using the formula:
- Assuming that population standard deviation and sample standard deviation are same:
SD_1 = SD_2 = σ = 4
- Then,
- The cohen's d can now be evaliated:
Cohen's d = 
Do you notice anything strange about those points ?
(0, 1),
(1, 2),
(2, 4),
(3, 8).
The y-coordinate of each point is (2) raised to the power of the x-coordinate.
First point: x=0, y=2⁰ = 1
Second point: x=1, y=2¹ = 2
Third point: x=2, y=2² = 4
Fourth point: x=3, y=2³ = 8
The equation of the curve appears to be
y = 2 ^ x .
So, after 10 hours, x=10, and y = 2¹⁰ = 1,024 .
Answer:
d. 16x² + y² = 16
Step-by-step explanation:
Standard form of an ellipse is:
x² / a² + y² / b² = 1
If b > a, the ellipse has a vertical orientation.
We can ignore options b and c, since they have minus signs in them.
Convert the other options to standard form:
a. x² / (1/16) + y² / (1/25) = 1
x² / (1/4)² + y² / (1/5)² = 1
1/4 > 1/5
d. x² + y² / 16 = 1
x² / 1² + y² / 4² = 1
1 < 4
e. x² / 16 + y² = 1
x² / 4² + y² / 1² = 1
4 > 1
Of these three options, only d is a vertical ellipse.
