Were the dollar amounts in the answer below rounded to the nearest dollar? Orginal problem was : Suppose that the Best Custom Calendar Company models the coming year's revenuse and cost functions, in thousands of dollars, to be R(x)=-x^3+9x^2 and C(x)=2x^3-12x^2+30x, where x represents units of 1000 calendars and where the model is thought to be accurate up to approximately x=5.5. Round the dollar amounts to the nearest dollar. What is the profitable zone of calendar production levels? What is the maximum profit and the level of production that will generate the maximum profit? What is the maximum loss and the level of production generating this maximum loss when the compnay is not in the profitable zone?
Minimization
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in terms of global or local range. The definition of minimization in the thesaurus is the process of reducing something to a small amount, value, or position. Minimization (noun) is an instance of belittling or disparagement.
Maxima and Minima
The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.
Derivatives
A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which one would be easier? Walking on the road would be much easier than climbing a mountain.
Concavity
In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the first derivative test and second derivative test to understand the concave behavior of the function.
Were the dollar amounts in the answer below rounded to the nearest dollar? Orginal problem was :
Suppose that the Best Custom Calendar Company models the coming year's revenuse and cost functions, in thousands of dollars, to be
R(x)=-x^3+9x^2 and C(x)=2x^3-12x^2+30x, where x represents units of 1000 calendars and where the model is thought to be accurate up to approximately x=5.5. Round the dollar amounts to the nearest dollar.
What is the profitable zone of calendar production levels?
What is the maximum profit and the level of production that will generate the maximum profit?
What is the maximum loss and the level of production generating this maximum loss when the compnay is not in the profitable zone?
ANSWER GIVEN:
R(x) =−x3+9x2R(x) =-x3+9x2 and C(x)=2x3−12x2+30xC(x)=2x3-12x2+30x
Here R(x) =−x3+9x2R(x) =-x3+9x2and C(x)=2x3−12x2+30xC(x)=2x3-12x2+30x
Profitable zone is when R(x)>C(x)R(x)>C(x)
−x3+9x2>2x3−12x2+30x-x3+9x2>2x3-12x2+30x
Solve the equality by simplifying it.
−3x3+21x2−30x>0-3x3+21x2-30x>0
3x3−21x2+30x<03x3-21x2+30x<0
x3−7x2+10x<0x3-7x2+10x<0
Factoring it, x(x−2)(x−5)<0x(x-2)(x-5)<0
The zeros are x=0,x=2,x=5x=0,x=2,x=5
The intervals are x<0x<0, 2<x<52<x<5, x>5x>5
By taking the test point the solution is 2<x<52<x<5
Thus the profitable zone is 2000<x<50002000<x<5000
P(x)=−3x3+21x2−30xP(x)=-3x3+21x2-30x
The maximum profit is find by derivative.
P'(x)=−9x2+42x−30P'(x)=-9x2+42x-30
Set P'(x)=0P'(x)=0
−9x2+42x−30=0-9x2+42x-30=0
3x2−14x+10=03x2-14x+10=0
Using quadratic formula, x=14±(−14)2−4(3)(10)√6x=14±(-14)2-4(3)(10)6
x=14±76√6x=14±766
x=7+19√3, x=7−19√3x=7+193, x=7-193
x=3.79, x=0.88x=3.79, x=0.88
To check for maximum point find second derivative
P''(x)=−18x+42P''(x)=-18x+42
P''(3.79)=−18(3.79)+42=−26.22P''(3.79)=-18(3.79)+42=-26.22
Therefore x=3.79x=3.79 is the maximum point.
The profit is maximum when 3790 units are produced.
The question says the model is accurate up to approximately x=5.5
For x>5x>5 it is not in profitable zone. So the maximum loss is at x=5.5x=5.5.
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