BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 3.1, Problem 66E
To determine

To find: The maximum profit per day and the amount of cans must be sold to obtain maximum profit.

Expert Solution

Answer to Problem 66E

The maximum profit per day is $450.

The number of cans that has to be sold to obtain the maximum profit is 1500.

Explanation of Solution

Given:

The function of profit (in dollars) to sell x cans of soda pop in one day is P(x)=0.001x2+3x1800.

Formula used:

Maximum or minimum value of a quadratic function:

The maximum or minimum value of a quadratic function f(x)=ax2+bx+c occurs at x=b2a.

If a>0, then the minimum value is f(b2a).

If a<0, then the maximum value is f(b2a).

Calculation:

The given quadratic function is, P(x)=0.001x2+3x1800.

Where, a=0.001, b=3 and c=1800.

From the above formula, it is known that maximum value exists at f(b2a) if a<0.

The quadratic function P(x)=0.001x2+3x1800 has the maximum value and it exists at f(32(0.001))=f(1500).

The number of cans that has to be sold to obtain the maximum profit is 1500.

Substitute x=1500 in P(x)=0.001x2+3x1800 as follows,

P(1500)=0.001(1500)2+3(1500)1800=2250+45001800=450

Maximum value of the function P(x)=0.001x2+3x1800 is 450.

Therefore, the maximum profit per day is $450 and the number of cans that has to be sold to obtain the maximum profit is 1500.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!