Items that can be produced and sold in order to make a profit is
[tex]5\leq n\leq 38[/tex]
Given :
The cost of producing n items is (950+63n). The items can be sold for (280-5n) per item
Cost price = [tex]950+63n[/tex]
Selling price is 280-5n for one item
Selling price of 'n' items
[tex]n \cdot (280-5n)\\280n-5n^2[/tex]
We know that , profit = selling price for n items - cost price for n items
We make profit only when selling price - cost price >=0
Lets replace the expressions
[tex]280n-5n^2 - (950+63n)\geq 0\\280n-5n^2 - 950-63n\geq 0\\\\-5n^2+217n-950\geq 0[/tex]
Solve the inequality for n
Apply the quadratic formula to solve for n
[tex]-5n^2+217n-950=0\\n=\frac{-217\pm \sqrt{217^2-4\left(-5\right)\left(-950\right)}}{2\left(-5\right)}\\n=\frac{-217\pm \:3\sqrt{3121}}{2\left(-5\right)}\\n=\frac{-217+3\sqrt{3121}}{2\left(-5\right)},\:n=\frac{-217-3\sqrt{3121}}{2\left(-5\right)}\\n=4.94022 ,\:n=38.45977[/tex]
Now we check the inequality
lets pick a number before 4.9 , n=4
[tex]-5n^2+217n-950\:\geq 0\\-162\geq 0 false\\\\Pick, n=10\\-5\left(10\right)^2+217\left(10\right)-950\geq 0\\720\geq 0 true\\\\Pick, n=40\\-5\left(40\right)^2+217\left(40\right)-950\geq 0\\-270\geq 0 false\\[/tex]
n=40 satisfies our inequality
40 lies between n=4.9 and n=38
So, [tex]5\leq n\leq 38[/tex] items can be produced and sold in order to make a profit
Learn more : brainly.com/question/15697949
