that for every 51 incre caps with the local baseball team's logo planning to increase the p consumer survey shows 10 as per year. What should the selling price bet R(x)=price # of items Leta be the number of $1 Increases in price, 270 Let R be the revenue, y>0. R(x) = (15+x) (600-30x) price. quantity R'(x)= (600-30x)+ (-30) (15+x) - 600-30x-450-30x = 150-60× = 30(5-2x) ut Ri (x)=0 0 = 30 (5-2x) 5-2x=0 x=2.5 → # of increases FDT D(x) Interval D'(a) sol'n Tord x425 + >0 x=25 N/A =0 max x>2.5 <0 - Sub x=2.5 into 15+x = price P= 15+ x = 15+25 = 175 .: the selling price should be $17.50 +0 maximize annual revenue Product of two positive numbers is 16. Find the two numbers such that the sum is minimum.
that for every 51 incre caps with the local baseball team's logo planning to increase the p consumer survey shows 10 as per year. What should the selling price bet R(x)=price # of items Leta be the number of $1 Increases in price, 270 Let R be the revenue, y>0. R(x) = (15+x) (600-30x) price. quantity R'(x)= (600-30x)+ (-30) (15+x) - 600-30x-450-30x = 150-60× = 30(5-2x) ut Ri (x)=0 0 = 30 (5-2x) 5-2x=0 x=2.5 → # of increases FDT D(x) Interval D'(a) sol'n Tord x425 + >0 x=25 N/A =0 max x>2.5 <0 - Sub x=2.5 into 15+x = price P= 15+ x = 15+25 = 175 .: the selling price should be $17.50 +0 maximize annual revenue Product of two positive numbers is 16. Find the two numbers such that the sum is minimum.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter4: Polynomial And Rational Functions
Section4.1: Quadratic Functions
Problem 103E
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this is a calculus optimization problem show all steps and include the chart of the first derivative test just like THIS EXAMPLE I'm showing you on paper.
Transcribed Image Text:that for every 51 incre
caps with the local baseball team's logo
planning to increase the p
consumer survey shows
10 as per year. What should the selling price bet
R(x)=price # of items
Leta be the number of $1 Increases in price, 270
Let R be the revenue, y>0.
R(x) = (15+x) (600-30x)
price.
quantity
R'(x)= (600-30x)+ (-30) (15+x)
- 600-30x-450-30x
= 150-60×
=
30(5-2x)
ut Ri (x)=0
0 = 30 (5-2x)
5-2x=0
x=2.5
→ # of
increases
FDT
D(x)
Interval
D'(a) sol'n
Tord
x425
+
>0
x=25
N/A =0
max
x>2.5
<0
-
Sub x=2.5 into 15+x = price
P= 15+ x
= 15+25
= 175
.: the selling
price should be
$17.50 +0
maximize annual
revenue
Transcribed Image Text:Product of two positive numbers is 16. Find the two numbers such that the sum is minimum.
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