is estimated at C (x) dollars, where x is the number of ties sold on a normal day and
c (x) = 0.0006 x ³ - 0.03 x ² +20 x +20. Find the value of x that will maximize the profit of the store.
Tip: L = R - C
Thanks!
c (x) = 0.0006 x ³ - 0.03 x ² +20 x +20. Find the value of x that will maximize the profit of the store.
Tip: L = R - C
Thanks!
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Each tie sells for $3.50, so if x ties are sold, the revenue (in dollars) is 3.5x
Profit = Revenue - Cost
P(x) = R(x) - C(x)
= 3.5x - (0.0006 x³ - 0.03 x² + 20x + 20)
= -0.0006x³ + 0.03x² - 16.5x - 20
dP/dx = -0.0018x² + 0.06x - 16.5
Set dP/dx = 0 and solve for x.
You must have a typo somewhere. dP/dx = 0 has no real solution. Are you sure about the coefficient of x in C(x)? Could it be 2.0? Or is the selling price wrong?
Profit = Revenue - Cost
P(x) = R(x) - C(x)
= 3.5x - (0.0006 x³ - 0.03 x² + 20x + 20)
= -0.0006x³ + 0.03x² - 16.5x - 20
dP/dx = -0.0018x² + 0.06x - 16.5
Set dP/dx = 0 and solve for x.
You must have a typo somewhere. dP/dx = 0 has no real solution. Are you sure about the coefficient of x in C(x)? Could it be 2.0? Or is the selling price wrong?