OQA 25

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course MTH 271

5/7/126:40pm

025. `query 25

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Question: `q Problem 1 b 7th edition 3.4.6 find two positive numbers such that the product is 192 and a sum of the first plus three times the second is a minimum.

What are the two desired numbers and how did you find them?

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Your solution:

the primary equation for this problem is: S = a + 3b

and a*b = 192 → b = 192/a

Replace b with the equation for b:

S = a + 3(192/a) → find the derivative to find the critical points:

S’ = 1 - 576/a^2

a is zero when equal to 24, -24, but the number is positive, so -24 is impossible.

Substitute a= 24 into the equation:

b = 192/24 = 8

so the two numbers are 24 and 8.

confidence rating #$&*:

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Given Solution:

`a First set up the primary equation S=x+3y (y being the 2nd number) and the secondary equation xy=192.

So S = x + 3(192/x).

We now maximize the function by finding critical points (points where the derivative is zero) and testing to see whether each gives a max, a min, or neither.

S ' = 1 - 576 / x^2, which is zero when x = sqrt(576) = 24 (or -24, but the problem asks for positive numbers).

For this value of x we get y = 192 / x = 192 / 24 = 8.

So the numbers are x = 24 and y = 8.

}Note that x = 24 does result in a min by the first derivative test, since S ' is negative for x < 24 and positive for x > 24. **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q Problem 5 80 apple trees in a certain field will yield an average of 400 per tree; each additional tree decreases the yield by 4 apples per tree. How many trees should be planted to maximize the yield?

the yield (per tree) function is (400 - 4x) and the number of trees is (80 + x).

the complete yield is equal to the number of trees times the yield per tree:

Y = (400 - 4x)(80 + x) → Y = -4x^2 + 80x + 32,000

and the first derivative is Y’ = -4x + 80 and the critical point of this function is x = 10

this should be the maximum value:

(80 + 10) = 90, (400- 4(10)) = 360

Y = 360*90 = 32,400

90 is the maximum number of trees.

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Given Solution:

`a If we let x stand for the number of trees added to the 80 then the yield per tree is 400 - 4 x, and there would be 80 + x trees.

The total yield is therefore

total yield = yield per tree * number of trees = (400 - 4 x)(80 + x) = -4 x^2 + 80 x + 32000

The derivative of this function is - 8 x + 80, and the second derivative is -4.

The derivative is zero when -8 x + 80 = 0, so the solution x = 10 is the critical value.

The second derivative is negative, so a graph of the function is concave down, indicating that the critical value is a maximum.

We conclude that the maximum yield is obtained by planting 20 additional trees, so that the total number of trees is 80 + 10 = 90..

The yield per tree will be 400 - 10 * 4 = 360 so the total yield will be 90 trees * 360 apples / tree = 32 400 apples.

If we evaluate the total-yield function -4 x^2 + 80 x + 32000 for x = 10, we get 32 400, verifying our solution.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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