Real World Quadratic
MAT222: Intermediate Algebra
June 23, 2014
The business world is very competitive and companies are in business to make money for example, convenience stores usually run displays to offer cheaper gas, cigarettes and some non-conventional items found in their stores. The use of the quadratic equations in the world would show the profitability of x number clerks on duty and plot the number of x clerks to show to ideal number to have on duty without wasting resources and producing a profit. The assigned problem is a manager must complete a graph to showing the daily profit of store.
Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
The first step is to set the equation in quadratic form.
To find the max amount of clerks needed to get the max amount of profit that day we will be using the formula (x= (-b) / (2(a))) which will also be part of the vertex.
Now, the problem can be solved to find the maximum profit by finding the value of x.
P=-25x2 +300x a=-25b=300
Plug our numbers into the appropriate place in out formula. We will then multiply 2 times -25 to get -50 then divide -300 by -50 getting 6. We find that the max amount of clerks needed to get the max amount of profits is 6 clerks. So the equation will look like this:
The maximum number of clerks to be profitable is six clerks should be on duty to increase their daily earning potential. Solve the problem by answering the maximum amount of profit. We will place values of x in the equation from the last solution.
P=-25x2 + 300x
P=-25(6) + 300(6)
P= -900 +1800
The graph would be a parabola going downward which would show the increase to decrease of profit based on...