The total cost, y naira, manufacturing x units of an article is given by the relation y=(5/4)x^2+20/x. Find the number of units of the articles for which the cost of manufacture is least and the corresponding least cost
To find the number of units of the article for which the cost of manufacture is least, we need to minimize the cost function y.
To do this, we'll take the derivative of y with respect to x and set it equal to zero:
dy/dx = (5/4)(2x) - 20/x^2 = (10/4)x - 20/x^2 = (5/2)x - 20/x^2
Setting this equal to zero and solving for x:
(5/2)x - 20/x^2 = 0
(5/2)x = 20/x^2
(x^3)(5/2) = 20
5x^3 = 40
x^3 = 8
x = 2
So the number of units of the article for which the cost of manufacture is least is 2.
To find the corresponding least cost, we substitute x = 2 into the cost function y:
y = (5/4)(2^2) + 20/(2)
y = (5/4)(4) + 10
y = 5 + 10
y = 15
Therefore, the number of units of the articles for which the cost of manufacture is least is 2, and the corresponding least cost is 15 naira.
To find the number of units of the article for which the cost of manufacture is the least, we need to minimize the total cost function y = (5/4)x^2 + 20/x.
To do this, we can take the derivative of the cost function with respect to x and set it equal to zero:
dy/dx = 0
Let's find the derivative:
dy/dx = (5/4)(2x) - 20/x^2
= (5/2)x - 20/x^2
Setting this derivative equal to zero, we get:
(5/2)x - 20/x^2 = 0
Multiplying through by x^2, we get:
(5/2)x^3 - 20 = 0
Dividing through by (5/2), we get:
x^3 - 8 = 0
Now, let's solve this equation:
x^3 = 8
Taking the cube root of both sides, we get:
x = 2
So, the number of units of the article for which the cost of manufacture is least is 2.
To find the corresponding least cost, we need to substitute this value of x back into the total cost function:
y = (5/4)(2^2) + 20/(2)
= (5/4)(4) + 10
= 5 + 10
= 15
Therefore, the corresponding least cost is 15 naira.