Question

the daily profit in dollars of a specialty cake shop is described by the function P(x)=-3x^2+168x-1920. Where x is the number of cakes prepared in one day. The maximum profit for the company occurs at the vertex of the parabola. How many cakes should be prepared per day in order to maximize profit?

Answer 1

Answer:

28

Step-by-step explanation:

We have the following function:

$f(x)=-3x^{2} +168x-1920$

where a=-3, b=168 and c=-1920

In order to calculate the maxium profit for the company, and how many cakes should be prepared in order to reach it, we have to calculate where the parabola's vertex is ubicated. To do so, we use the following formula:

$Xv= \frac{-b}{2a}$

$Xv= \frac{-168}{2.(-3)}$

$Xv= \frac{-168}{2.(-3)}$

$Xv= 28$

So 28 cakes should be prepared per day in order to maximize profit.

Answer 2

The number of cakes that should be prepared per day to maximize profit, P(x) is: 28.

From the question:

The profit function is given as;

• P(x)= -3x^2+168x-1920

In essence, the point of maximum profit is at the point where the parabola vertex is ubicated and can be evaluated as follows;

• x = -b/2a

in which case; a = -3, b = 168 and c = -1920

Therefore,

• x = -(168)/(2×-3)

• x = -168/-6

• x = 28

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