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Katena32 [7]
3 years ago
11

I would like to create a rectangular vegetable patch. The fencing for the east and west sides costs $4 per foot, and the fencing

for the north and south sides costs only $2 per foot. I have a budget of $128 for the project. What are the dimensions of the vegetable patch with the largest area I can enclose?
Mathematics
1 answer:
Anestetic [448]3 years ago
5 0
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y.  the dimensions you want are therefore x and y.

The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y.  We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.



The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128.  Let's reduce this by dividing all three terms by 4:  2x + y = 32.

Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32.  The formula for area is A = L * W.  Solving 2x + y = 32 for y, we get y = -2x + 32.

We can now eliminate y.  The area of the patch is (x)(-2x+32) = A.  We want to maximize A.

If you're in algebra, find the x-coordinate of the vertex of this quadratic equation.  Remember the formula x = -b/(2a)?  Once you have calculated this x, subst. your value into the formula for y:  y= -2x + 32.

Now multiply together your x and y values to obtain the max area of the patch.


If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero.  This approach should give you the same x value as before; the corresponding y value will be the same;  y=-2x+32.

Multiply x and y together.  That'll give you the maximum possible area of the garden patch.
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Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846

If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours.

This is 1 subtracted by the pvalue of Z when X = 8.1. So

Z = \frac{X - \mu}{\sigma}

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Z = \frac{X - \mu}{s}

Z = \frac{8.1 - 8.4}{0.2846}

Z = -1.05

Z = -1.05 has a pvalue of 0.1469

1 - 0.1469 = 0.8531

85.31% probability that their mean rebuild time exceeds 8.1 hours.

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