AsmalltieshopsellstiesforGHC3.50each.Thedailycostfunctionisestimatedto beC(x)dollarswherexisthenumberoftiessoldonatypicaldayandC(x)= 0.0006x3–0.03x2+2x+20.Findthevalueofxthatwillmaximizethestore’sdaily profit

To find the value of x that will maximize the store's daily profit, we first need to find the profit function. The profit function is the revenue function minus the cost function.

1. The revenue function is the price per tie times the number of ties sold, or R(x) = 3.50x.

2. The cost function is given as C(x) = 0.0006x^3 - 0.03x^2 + 2x + 20.

3. The profit function P(x) is therefore P(x) = R(x) - C(x) = 3.50x - (0.0006x^3 - 0.03x^2 + 2x + 20).

Simplify the profit function to get P(x) = -0.0006x^3 + 0.03x^2 + 1.5x - 20.

4. To find the maximum profit, we need to find the derivative of the profit function and set it equal to zero.

The derivative of P(x) is P'(x) = -0.0018x^2 + 0.06x + 1.5.

5. Set P'(x) = 0 and solve for x. This will give you the x-values where the profit function has a maximum or minimum.

6. To determine whether these x-values are maximums or minimums, you can use the second derivative test. The second derivative of P(x) is P''(x) = -0.0036x + 0.06.

7. If P''(x) is less than zero for a given x-value, then that x-value is a maximum. If P''(x) is greater than zero, then that x-value is a minimum.

8. Solve P''(x) < 0 to find the x-values that give a maximum profit.

Please note that the above steps involve calculus. If you are not familiar with calculus, you may need to consult a textbook or online resource to understand the steps.