In making electric kettles, the manufacturer performs a cost control study and discovers that to produce x kettles the cost per kettle c(x) is given by c(x) =4lnx + (3-10/x)^2 hundred dollars with a minimum production capacity per day of 10 kettles. How many kettles should be purchased to keep the cost per kettle a minimum?
To find the number of kettles that should be purchased to keep the cost per kettle at a minimum, we need to find the value of x that minimizes the cost function c(x) = 4lnx + (3 - 10/x)^2.
To do this, we can take the derivative of c(x) with respect to x, and set it equal to zero to find the critical points:
c'(x) = 4/x + 2(3 - 10/x)(-10/x^2)
Setting c'(x) equal to zero:
0 = 4/x + 2(3 - 10/x)(-10/x^2)
Simplifying the equation:
0 = 4/x - 20(3 - 10/x)/x^2
Multiplying through by x^2 to clear the fractions:
0 = 4x - 20(3x - 10)/x
Expanding and simplifying further:
0 = 4x - 60 + 200/x
Combining like terms:
60 = 4x + 200/x
Rearranging the equation:
4x + 200/x - 60 = 0
To solve this equation, we can make a substitution:
Let u = x/10
Then, the equation becomes:
4u + 20/u - 6 = 0
Multiplying through by u:
4u^2 + 20 - 6u = 0
Rearranging and factoring:
4u^2 - 6u + 20 = 0
2u^2 - 3u + 10 = 0
Using the quadratic formula, we find:
u = (3 ± sqrt((-3)^2 - 4(2)(10))) / (2(2))
u = (3 ± sqrt(9 - 80)) / 4
u = (3 ± sqrt(-71)) / 4
Since the discriminant is negative, there are no real solutions for u, and hence, no real solutions for x.
However, the given problem states that there is a minimum production capacity of 10 kettles per day. Therefore, we can conclude that the cost per kettle will be minimized when the company purchases 10 kettles.
To find the number of kettles that should be purchased to keep the cost per kettle at a minimum, we need to determine the value of x that minimizes the cost function c(x) = 4lnx + (3 - 10/x)^2.
To do this, we can take the derivative of the cost function with respect to x and set it equal to zero.
The derivative of the cost function c(x) with respect to x is given by:
c'(x) = (4/x) + 2(3 - 10/x)(10/x^2)
To find the minimum, we set c'(x) equal to zero and solve for x:
(4/x) + 2(3 - 10/x)(10/x^2) = 0
Multiplying through by x^2 to eliminate the denominators:
4x + 2(3x^2 - 10) = 0
Expanding and rearranging the equation:
6x^2 - 20x + 6 = 0
Dividing through by 2 to simplify the equation:
3x^2 - 10x + 3 = 0
Now, we can solve this quadratic equation to find the value of x that minimizes the cost per kettle. We can do this by factoring, completing the square, or using the quadratic formula.
However, upon inspecting the equation, it does not appear to factor nicely. Therefore, we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from the quadratic equation:
x = (-(-10) ± √((-10)^2 - 4(3)(3))) / 2(3)
= (10 ± √(100 - 36)) / 6
= (10 ± √64) / 6
Simplifying under the square root:
√64 = 8
So the two possible solutions for x are:
x1 = (10 + 8) / 6 = 18 / 6 = 3
x2 = (10 - 8) / 6 = 2 / 6 = 1/3
Since the minimum production capacity per day is 10 kettles, we can discard the option x2 = 1/3 as it is less than the minimum requirement.
Therefore, the number of kettles that should be purchased to keep the cost per kettle at a minimum is x = 3.
c(x) = 4lnx + (3 - 10/x)^2
dc/dx = 4/x + 2(3 - 10/x)(10/x^2)
so, where is dc/dx = 0?
Is that a max or a min?
Does it satisfy the imposed conditions?