Help me Please I have a test tomorrow!

Question

The profit from a business is described by the function P(x) = –x² + 10x – 4, where x is the number of items made, in thousands, and P(x) is the profit in dollars. How many items will maximize the profit.

Answer 1

For a quadratic equation, the vertex is found where x = -b/2a

The equations is -x² +10x - 4, so a is -1, b is10, and c is -4

so -b/2a is -10/2(-1) = -10/-2 = 5

If you use Desmos Graphing Calculator to visualize, you'll see that the curve is indeed at a maximum where x = 5 (5,000 units). This yields a profit of 21. Not specified if profit is in 000's as well, but it can probably be assumed.

Answer 2

Since you indicated Algebra 1, I will show you how to solve this problem without using Calculus.

P(x) = –x² + 10x – 4 = - ( x² - 10x + 4 ) = - ( x² - 10x + 25 - 25 + 4 ) = - ( (x-5)² - 21 ) = 21 - (x-5)²

I used the formula (a - b)² = a² - 2ab + b²

(x-5)² ≥ 0, therefore P(x) is largest when x - 5 = 0 or

x = 5

In case the question was mislabeled, a solution with Calculus is below

P(x) = –x² + 10x – 4

Let us take the derivative of P(x)

P’(x) = - 2 x + 10

The value of P(x) is maximized, when P’(x) = 0

x = 5

Answer 3

P(x) =-x^2 +10x -4

= -(x^2 -10x) -4

= -(x^2 -10x +25) -4 +25

= -(x-5)^2 +21 in vertex form where (5,21) is the vertex = max point. 21 thousand dollars = maximum profit

x=5 thousand

P(5) = -5^2 +10(5) -4 = 50-25-4 = 21 thousand dollars

50,000-25000 -4000 = $21,000 profit

or use calculus

take the derivative and set equal to zero

P'(x) = -2x +10 = 0

x =10/2= 5