The larger number is 99. The smaller number is 21.
Solve the following two problems using Calculus. Show all of your work on a separate sheet of paper, making clear how you arrive
d at all your solutions. Please be aware that you are using similar thinking for both of these problems, but you are solving for maximum area in Part A and minimum cost in Part B. Enjoy
A. If the materials for the fence cost $12 per foot, find the dimensions of the
corral for the largest possible area that can be enclosed with $2400 worth of fence.
B. If he only requires an area of 288 square feet for his vegetable garden, find the minimum cost of putting up this fence, if the cost per foot is $10. Indicate the dimensions of the garden that will minimize the cost.
A. If the materials for the fence cost $12 per foot, find the dimensions of the
corral for the largest possible area that can be enclosed with $2400 worth of fence.
B. If he only requires an area of 288 square feet for his vegetable garden, find the minimum cost of putting up this fence, if the cost per foot is $10. Indicate the dimensions of the garden that will minimize the cost.
1 answer:
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Answer:
<h2>Diana uses 5/8 square yard of fabric for both pillows.</h2><h2>There remains 3/8 square yard of fabric.</h2>Step-by-step explanation:
Givens
- The total amount of fabric is 1 square yard.
- One pillow requires 3/8 square yard of fabric.
- A second pillow requires 2/8 square yard of fabric.
To find the total amount of fabric Diana used, we must sum those fractions
Diana uses 5/8 square yard of fabric for both pillows.
Now, we subtract to find the remaining amount of fabric
Therefore, there remains 3/8 square yard of fabric.
Answer:
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.
Step-by-step explanation:
We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.
Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.
Let X = <u><em>back-to-college family spending on electronics</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean family spending = $237
= standard deviation = $54
(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)
P(X < $150) = P( <
) = P(Z < -1.61) = 1 - P(Z
1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.
(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)
P(X > $390) = P( >
) = P(Z > 2.83) = 1 - P(Z
2.83)
= 1 - 0.9977 = <u>0.0023</u>
The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.
(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)
P($120 < X < $175) = P(X < $175) - P(X $120)
P(X < $175) = P( <
) = P(Z < -1.15) = 1 - P(Z
1.15)
= 1 - 0.8749 = 0.1251
P(X < $120) = P( <
) = P(Z < -2.17) = 1 - P(Z
2.17)
= 1 - 0.9850 = 0.015
The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.
Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>