Mathematical Modeling with Quadratic Functions: Maximum and Minimum Problem???

Question

The marketing department at Texas Instruments has found that, when certain calculators are sold at a price of "p" dollars per unit, the number "x" of calculators sold is given by the demand equation:
x = 21000 - 150p

1. Express the revenue "R" as a function of the price "p"
2. What unit price should be used to maximize revenue? (Complete the square of your revenue function to identify the vertex)
3.If this price is charged, what is the maximum revenue?
4. How many units are sold at this price?

I then haveto graph this, I'm having troubles understanding this concept and don't know where to start to set up the revenue as a function of the price equation??Help please??

Answer 1

1. This is easy; just realize that revenue equals price per unit times units sold, then replace x (units sold) with the expression in terms of p.

R = px
R = p(21000 - 150p)
R = 21000p - 150p²

2. Ok, entirely not sure how completing the square gives the vertex, but this is how you complete the square:

-150p² + 21000p = 0
p² - 140p = 0
p² - 140p + 70² = 70²
(p - 70)² = 70²
p - 70 = ±70
p = 70 ± 70

I guess that means 70 is the vertex? Anyway, the derivative method I use next gives that answer, and I trust it more because I am familiar with it...

Anyway, the way I know to maximize revenue is by a derivative. The zero of the derivative is a local maximum or minimum, so find the zero and that is the price of maximum revenue:

R' = 21000 - 300p
0 = 21000 - 300p
p = 21000/300
p = $70

3. Easy substitution into the revenue function; just plug in p and solve:

R = 21000p - 150p²
R = 21000(70) - 150(70²)
R = $735,000

4. Also easy substitution; plug p into the equation for x and solve:

x = 21000 - 150p
x = 21000 - 150(70)
x = 10,500

Answer 2

x = 21000 - 150p
Revenue is product of calculators (x) sold and unit price of each
-------------------------------------
R = x * p = p*(21000 - 150p) = 21000 p - 150 p^2
its R as a function of p
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find that x for which R is maximum
R = - 150 p^2 + 21000 p
R = - 150[p^2 - 140 p + 4900 - 4900]
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making square: ax^2+bx+c >>> add/subtract c'= b^2/4a
b=140, a=1 so c' 140*140/4 = 4900 = (70)^2
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R = - 150[p -70)^2 - 4900]
R = 150 [(70)^2 - (p -70)^2] ----(1)
R will be maximum when (p-70)^2 (which is reducing R) is zero
so p-70 = 0 >>>> p =70 price for R (max)
---------------------
at this price p=70
Rmax = 150 [(70)^2 - (p -70)^2]
= 150 [ 4900 - 0] = 735000
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at this price. units sold
x = 21000 - 150*70 = 10500 units sold

hope it helps you
These are very necessary for dealings in any field, and later when you handle business or position of responsibility.

Answer 3

hi, you will discover the optimal/minimal of a quadratic equation by the assumption of differentiation. by the assumption of MAXIMA $ MINIMA we get the assumption . f(x)=x^2+8x+10 f'(x)=2x+eight f''(x)=2>0 by double differentiatiate we come to be attentive to that the function has minimal at f'(x)=0 2x+eight=0 (As f'(x)=0 provides the optimal or minimal fee) x= --4. So we detect f(--4)=(--4)^2+eight(--4)+10 =sixteen--32+10 = --6 consequently the minimal fee of the function is --6.