1) The total cost "C" for ordering and storing x units is C(x) = 2x + (300000)/(x). What order size will produce a minimum cost?
2) Find two positive numbers where the second number is the reciprocal of the first and their sum is minimum.
3) A dairy farmer plans to fence in a rectangular pasture adjacent toa river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
P.S. If you guys can, can you guys please show steps? I really want to learn this.
Methods allowed to use are the mean value theorem, closed interval method (to find the global max and min), and the derivative test to find the extremes.
2) Find two positive numbers where the second number is the reciprocal of the first and their sum is minimum.
3) A dairy farmer plans to fence in a rectangular pasture adjacent toa river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
P.S. If you guys can, can you guys please show steps? I really want to learn this.
Methods allowed to use are the mean value theorem, closed interval method (to find the global max and min), and the derivative test to find the extremes.
-
1.
Find the derivative of C(x). C'(x) = 2 - 300 000/x².
Now set the derivative to equal 0: 2 - 300 000/x² = 0 → 2 = 300 000/x² → x² = 150 000 → x = ±√150 000. Reject -√150 000 because you can't order a negative number of units
Test one lower and one higher number than √150 000 to confirm that there is a minimum at this point (it will be). Therefore, since x = √150 000 is a minimum, ordering √150 000 units will give you a minimum cost.
2.
x + 1/x = y is the equation. Find the derivative of the equation: y' = 1 - 1/x²
Set the derivative to equal 0 → 1 - 1/x² = 0 → x² = 1 → x = ±1, however x must be a positive number so you can reject x = -1.
Again test a number lower and higher than 1 to make sure there is a minimum at x = 1 (again it will be). Therefore, the first number is 1 and second number is 1/1 = 1.
3.
y = 2x + 180 000/x. In this equation y is the amount of fencing needed and x is the amount of fencing long one of the sides perpendicular to the river and z is the amount of fencing parallel to the river.
~~river~~
.|------------ |
x |-------------|
.| _______|
z
xz = 180 000 → z = 180 000/x. Therefore the amount of fencing needed is y = 2x + z = 2x + 180 000/x.
Therefore, now that we have the equation we just follow the same steps as in the previous problems.
Find the derivative and set it to equal 0: y' = 2 - 180 000/x² = 0 → x² = 90 000 → x = √90 000.
Test a number lower and higher to make sure = √90 000 is a minimum. If it is then you can plug this number back into the original equation: y = 2x + 180 000/x to find the minimum amount of fencing needed.
I hope this helps :)
Find the derivative of C(x). C'(x) = 2 - 300 000/x².
Now set the derivative to equal 0: 2 - 300 000/x² = 0 → 2 = 300 000/x² → x² = 150 000 → x = ±√150 000. Reject -√150 000 because you can't order a negative number of units
Test one lower and one higher number than √150 000 to confirm that there is a minimum at this point (it will be). Therefore, since x = √150 000 is a minimum, ordering √150 000 units will give you a minimum cost.
2.
x + 1/x = y is the equation. Find the derivative of the equation: y' = 1 - 1/x²
Set the derivative to equal 0 → 1 - 1/x² = 0 → x² = 1 → x = ±1, however x must be a positive number so you can reject x = -1.
Again test a number lower and higher than 1 to make sure there is a minimum at x = 1 (again it will be). Therefore, the first number is 1 and second number is 1/1 = 1.
3.
y = 2x + 180 000/x. In this equation y is the amount of fencing needed and x is the amount of fencing long one of the sides perpendicular to the river and z is the amount of fencing parallel to the river.
~~river~~
.|------------ |
x |-------------|
.| _______|
z
xz = 180 000 → z = 180 000/x. Therefore the amount of fencing needed is y = 2x + z = 2x + 180 000/x.
Therefore, now that we have the equation we just follow the same steps as in the previous problems.
Find the derivative and set it to equal 0: y' = 2 - 180 000/x² = 0 → x² = 90 000 → x = √90 000.
Test a number lower and higher to make sure = √90 000 is a minimum. If it is then you can plug this number back into the original equation: y = 2x + 180 000/x to find the minimum amount of fencing needed.
I hope this helps :)