3(x +6x) ,ux2-1er -21-0 5. Solve for x, 16x2-16x-21=0 6, Give the f( x) = 3x2-4x
ID: 2888494 • Letter: 3
Question
3(x +6x) ,ux2-1er -21-0 5. Solve for x, 16x2-16x-21=0 6, Give the f( x) = 3x2-4x + 8 , find the difference of quotient 7.) The demand equation ABC Company for certain brand of videocassette is given by p d(x)-- 0.01xx2- .2x + 8 where p is the wholesale unit price in $ and x is the quantity demand each week measure in units of a thousand. a) What is the maximum quantity demand per week b) if the corresponding supply function is given by p s(x) 01x2 +1x +3 find the equilibrium quantity and price. 8. Graph the equation 3/8 x2 -2x+2 9. Find the Equilibrium quantity and Equilibrium price. 11P + 3X-66-0 and 2p2 + p-X= 10 10. Suppose puritron, a manufacture of water filter Has a monthly fixed cost of 1 2 + 10 X and the total Revenue realized 1x given by total Revenue-0.0005 20 X Find the total profit function, what is the profit when the total of production is 10000 filter/ month? 11) The price p and the quantity x units sold of certain product obey the demand function p =-1/6X4 100 express the r as a function of x a) what is r if 200 units are sold b) what is quantity of x that maximized r c) what is the maximum r d) what price should the company change to maximize r? 13. The marketing department at Texas Instrument has found that, when certain calculators are sold at certain price of $P per unit, the revenue R as a function of price p is -150P2 -12000P what unit price should be established in order to maximize revenue? If this price is changed, what is the maximum revenue?Explanation / Answer
multiple questions posted.please post each question seperately
5)
16x2-16x-21=0
=>16x2-16x=21
=>x2-x=(21/16)
=>x2-2*(1/2)x+(1/2)2=(21/16)+(1/2)2
=>(x-(1/2))2=(21/16)+(1/4)
=>(x-(1/2))2=(25/16)
=>(x-(1/2))2=(5/4)2
=>(x-(1/2))=-(5/4),(x-(1/2))=(5/4)
=>x=-(5/4)+(1/2),x=(5/4)+(1/2)
=>x=-(3/4),x=(7/4)
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7)
given demand,p=d(x) =-0.01x2-0.2x +8
(a)
for maximum quantity demand ,p=0
=>-0.01x2-0.2x +8=0
=>0.01x2+0.2x =8
=>0.01(x2+20x) =8
=>(x2+20x) =800
=>x2+20x-800=0
=>x2+40x-20x-800=0
=>(x+40)(x-20)=0
=>x-20=0
=>x=20
maximum quantity demand per week is 20 thousand units
(b)
at equilibrium ,p=d(x)=s(x)
=>-0.01x2-0.2x +8 = 0.01x2+0.1x +3
=>0.01x2+0.2x +0.01x2+0.1x = 8 - 3
=>0.02x2+0.3x = 5
=>2x2+30x = 500
=>2x2+30x - 500=0
=>2(x2+15x - 250)=0
=>(x2+15x - 250)=0
=>(x+25)(x-10)=0
=>x-10=0
=>x=10
p=d(10)
=>p=-0.01(10)2-0.2(10) +8
=>p=-1-2 +8
=>p=5
equilibrium quantity is 10 thousand units
equilibrium price is 5 dollars per unit
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