The monthly revenue R (in thousands of dollars) from the sales of a digital picture frame is approximated by R(p) = −10p^2 + 1480p, where p is the price per unit (in dollars).
a) Find the unit price that will yield a maximum monthly revenue.
answer: -b/2a=-(1480/2(-10))=7400
is this answer correct??
b) What is the maximum monthly revenue?
1480/20 is greater than 1480? Typo alert!
max value is c - b^2/4a
a) To find the unit price that will yield a maximum monthly revenue, we can use the vertex formula.
The equation for the revenue is R(p) = -10p^2 + 1480p.
We can rewrite this equation in the form R(p) = ap^2 + bp, where a = -10 and b = 1480.
The vertex formula is given by p = -b/2a.
Substituting the values, we get p = -(1480) / (2 * (-10)) = 74.
Therefore, the unit price that will yield a maximum monthly revenue is $74.
b) To find the maximum monthly revenue, we can substitute the value of p = 74 into the equation R(p) = -10p^2 + 1480p:
R(74) = -10(74)^2 + 1480(74)
= -10(5476) + 109520
= -54760 + 109520
= 54760.
Therefore, the maximum monthly revenue is $54,760.
To find the unit price that will yield a maximum monthly revenue, we need to find the vertex of the parabola represented by the revenue function R(p) = -10p^2 + 1480p.
a) The formula for finding the x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c is given by:
x = -b / (2a)
In this case, the coefficient of the p^2 term is a = -10 and the coefficient of the p term is b = 1480. Plugging these values into the formula, we get:
x = -(1480) / (2(-10))
x = -7400 / (-20)
x = 370
Therefore, the unit price that will yield a maximum monthly revenue is $370.
b) To find the maximum monthly revenue, we substitute the value of the unit price (p) into the revenue function R(p):
R(370) = -10(370)^2 + 1480(370)
R(370) = -10(136900) + 547600
R(370) = -1369000 + 547600
R(370) = 410600
Therefore, the maximum monthly revenue is $410,600.