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Elodia [21]
3 years ago
12

S= zh - 2zt^3 solve for z

Mathematics
2 answers:
Tresset [83]3 years ago
8 0

Answer:

z = \frac{s}{h-2t^3}

Step-by-step explanation:

Given

s = zh - 2zt³ ← factor out z from each term

s = z(h - 2t³ ) ← divide both sides by h - 2t³

\frac{s}{h-2t^3} = z

zysi [14]3 years ago
6 0

Hi ;-)

s=zh-2zt^3\\\\z(h-2t^3)=s \ \ /:(h-2t^3)\\\\z=\dfrac{s}{h-2t^3}

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A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are
skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20​),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20​)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

c)

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20​)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0
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