Need help figuring this out step by step attached is a formula sheet that you can use to follow *TATTOO"CHEAT SHEET - USE OFTEN!youRTHE BASICSderv.PX) MPCO) MCRX) MRX→ fcx)+y (ly CAN BE +,Ø,-)X f'x) SLOPE(SLOPE CAN BE +,Ø,integ.- {0 = MAx OR MIN OR H.P.I.3)X *f"(x) + CONCAVITY (CONCAVITY IS ,A,Ø { Bis POINT OF INFLECTION})DERIVATIVESPRODUCTS AND QUOTIENTSu isAFUNCTION,X IS VAR.,e AND n y= u y'= u'v -uv'y=x^y'= nxn-iU AND Vyouv y'- u'v+uv'%3Dn-Iy'= nu" (u')y=e" y'- u'e"y=Lnu y'= 4 doutdiny=u^AREFUNCTIONSCONSTANTSLOGS AND EXPONENTSy=a" y'=a"u'ına ) WHERE Uy-a* y = a*x' In a fa is const,Fa*(1) inaINTEGRALSIS A FUNC,x^dx+K, n+-IX IS VARVABLEntldx → U'+k, n+-1Un(e*) =x (SIMPLIFIED, NOT DERIVATI VE )=X (SIMPUFIED, NOT DERIVATIVE)y= Ju'e"dr → e" + kin(MN) = Ln M+ inNin (A) - LnM -UnNin (M) = PlnMy= logax ay =xWHERE M» In u +K n=-lEN ARESTEPSFUNCTIONS.(CONVERSIONSNOT DERIVATIVĖS1) MAKE IT PRETTY. WHICH INTEGRAL?2) FIND U; CREATE U'3) WE HAVE4) MAKE IT LOOK LIKE TEMPLATE5) PERFORM INTEGRALWE WANT.CHANGE OF BASE:y=loga xlogaDEFINITE INTEGRALSlog XIn X%3DALSOUnay=Jax^dx = afx°dxScan"y= J(ax^+bx") dxSFondh Fo = Fb) - Fla)Jfandk=Fx)%3D%3Da 9. If the Marginal Cost for a product is found to be MC =1500and the cost of 80 units is found to be15x+100'$6000, what are:a. the Total Costs?b. the Fixed Costs?

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Description: Need help figuring this out step by step attached is a formula sheet that you can use to follow *TATTOO"CHEAT SHEET - USE OFTEN!youRTHE BASICSderv.PX) MPCO) MCRX) MRX→ fcx)+y (ly CAN BE +,Ø,-)X f'x) SLOPE(SLOPE CAN BE +,Ø,integ.- {0 = MAx OR MIN OR H

Given, Marginal cost for the a product , MC¯ = 150015x + 100 The cost of 80…

Answered: 9. If the Marginal Cost for a product…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Need help figuring this out step by step attached is a formula sheet that you can use to follow

*TATTOO"CHEAT SHEET - USE OFTEN!
youR
THE BASICS
derv.
PX) MP
CO) MC
RX) MR
X→ fcx)+y (ly CAN BE +,Ø,-)
X f'x) SLOPE(SLOPE CAN BE +,Ø,
integ.
- {0 = MAx OR MIN OR H.P.I.3)
X *f"(x) + CONCAVITY (CONCAVITY IS ,A,Ø { Bis POINT OF INFLECTION})
DERIVATIVES
PRODUCTS AND QUOTIENTS
u isA
FUNCTION,
X IS VAR.,
e AND n y= u y'= u'v -uv'
y=x^
y'= nxn-i
U AND V
youv y'- u'v+uv'
%3D
n-I
y'= nu" (u')
y=e" y'- u'e"
y=Lnu y'= 4 doutdin
y=u^
ARE
FUNCTIONS
CONSTANTS
LOGS AND EXPONENTS
y=a" y'=a"u'ına ) WHERE U
y-a* y = a*x' In a fa is const,
Fa*(1) ina
INTEGRALS
IS A FUNC,
x^dx
+K, n+-I
X IS VARVABLE
ntl
dx → U'
+k, n+-1
Un(e*) =x (SIMPLIFIED, NOT DERIVATI VE )
=X (SIMPUFIED, NOT DERIVATIVE)
y= Ju'e"dr → e" + k
in(MN) = Ln M+ inN
in (A) - LnM -UnN
in (M) = PlnM
y= logax ay =x
WHERE M
» In u +K n=-l
EN ARE
STEPS
FUNCTIONS.
(CONVERSIONS
NOT DERIVATIVĖS
1) MAKE IT PRETTY. WHICH INTEGRAL?
2) FIND U; CREATE U'
3) WE HAVE
4) MAKE IT LOOK LIKE TEMPLATE
5) PERFORM INTEGRAL
WE WANT.
CHANGE OF BASE:
y=loga x
loga
DEFINITE INTEGRALS
log X
In X
%3D
ALSO
Una
y=Jax^dx = afx°dx
Scan"
y= J(ax^+bx") dx
SFondh Fo = Fb) - Fla)
Jfandk=Fx)
%3D
%3D
a

9. If the Marginal Cost for a product is found to be MC =
1500
and the cost of 80 units is found to be
15x+100'
$6000, what are:
a. the Total Costs?
b. the Fixed Costs?

Expert Solution

Step 1

$G i v e n, M a r g i n a l \cos t f o r t h e a p r o d u c t, \bar{M C} = \frac{1500}{15 x + 100} T h e \cos t o f 80 u n i t s i s f o u n d t o b e $ 6000 .$

Step 2

$P a r t a N o w, \bar{M C} = \frac{1500}{15 x + 100} \Rightarrow \frac{d C (x)}{d x} = \frac{1500}{15 x + 100} w h e r e C (x) i s t h e t o t a l \cos t \Rightarrow d C (x) = \frac{1500}{15 x + 100} d x I n t r e g r a t i n g o n b o t h s i d e s, w e g e t \Rightarrow \int d C (x) = \int \frac{1500}{15 x + 100} d x \Rightarrow C (x) = \int \frac{1500}{15 x + 100} d x L e t 15 x + 100 = t t h e n 15 d x = d t \Rightarrow C (x) = \int \frac{1500}{t} \cdot \frac{1}{15} d t \Rightarrow C (x) = \int \frac{100}{t} d t \Rightarrow C (x) = 100 \int \frac{1}{t} d t \Rightarrow C (x) = 100 \ln (t) + K w h e r e k i s c o n s \tan t \Rightarrow C (x) = 100 \ln (15 x + 100) + K S i n c e, t h e \cos t o f 80 u n i t s i s $ 6000, \Rightarrow C (80) = 6000 \Rightarrow 100 \ln (15 (80) + 100) + K = 6000 \Rightarrow 100 \ln (1200 + 100) + K = 6000 \Rightarrow 100 \ln (1300) + K = 6000 \Rightarrow K = 6000 - 100 \ln (1300) \Rightarrow K = 5282.98804 \Rightarrow K \approx 5283 S o, t h e t o t a l \cos t i s C (x) = 100 \ln (15 x + 100) + 5283$

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