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3 years ago

9

A manufacturer cuts squares from the corners of a rectangular plece of sheet metal that measures 2 inches by 7 inches (see Figur

e 1). The manufacturer then
folds the metal upward to make an open-topped box (see Figure 2). Letting x represent the side-lengths (in inches) of the squares, use the ALEKS graphing
calculator to find the value of x that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

The value of x that maximizes the volume enclosed by this box is 0.46 inches

The maximum volume is 3.02 cubic inches

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The volume of the open-topped box is equal to

$V=LWH$

where

$L=(7-2x)\ in\\W=(2-2x)\ in\\H=x\ in$

substitute

$V=(7-2x)(2-2x)x$

Convert to expanded form

$V=(7-2x)(2-2x)x\\V=(14-14x-4x+4x^{2})x\\V=14x-14x^2-4x^2+4x^{3}\\V=4x^{3}-18x^{2} +14x$

using a graphing tool

Graph the cubic equation

Remember that

The domain for x is the interval -----> (0,1)

Because

If x>1

then

the width is negative (W=2-2x)

so

The maximum is the point (0.46,3.02)

see the attached figure

therefore

The value of x that maximizes the volume enclosed by this box is 0.46 inches

The maximum volume is 3.02 cubic inches

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Zanzabum

Answer:

a) Null hypothesis: $\mu \leq 40000$

Alternative hypothesis: $\mu > 40000$

b) $t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$

c) The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since the sample size is large enough we cna use the z distribution as an approximation for the statsitic on this case.

Since is a one right tailed test the p value would be:

$p_v =P(t_{(1699)}>2.438)=0.0074$

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance. So we can conclude that the true mean is higher than 40000 at the significance level assumed.

Step-by-step explanation:

Data given and notation

$\bar X=41182$ represent the sample mean

$s=19990$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =40000$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is greater than 40000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 40000$

Alternative hypothesis: $\mu > 40000$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$

Part c: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since the sample size is large enough we cna use the z distribution as an approximation for the statsitic on this case.

Since is a one right tailed test the p value would be:

$p_v =P(t_{(1699)}>2.438)=0.0074$

Part d: Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance. So we can conclude that the true mean is higher than 40000 at the significance level assumed.

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