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For a particular company, the profit P generated by selling
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For a particular company, the profit P generated by selling Q units of a certain product is given by the formula \(P = 128 + (\frac{Q^2}{4} + 4Q  16)^z\), where z > 0. The maximum profit is achieved when Q = (A) 2 (B) 4 (C) 8 (D) 16 (E) 32
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Originally posted by emmak on 08 May 2013, 22:15.
Last edited by Zarrolou on 09 May 2013, 09:07, edited 2 times in total.
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Re: For a particular company, the profit P generated by selling
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08 May 2013, 22:48
emmak wrote: For a particular company, the profit P generated by selling Q units of a certain product is given by the formula P = 128 + (–Q2/4 + 4Q – 16)^z, where z > 0. The maximum profit is achieved when Q = (A) 2 (B) 4 (C) 8 (D) 16 (E) 32 The answer will be [C], but I believe it will hold for only odd values of z. The above expression can be reduced to P = 128 + {[Q^2 16Q + 64]/2}^z *(1)^z => 128+ [Q8/2]^2z * (1)^z for all values of z, [Q8/2]^2z will be an increasing function with its minima at 8. Also, for odd values of z (1)^z = 1, which will be reversing the sign of the same. Hence, for Q=8 and z is odd, the above expression will have a maxima. I am still unsure as to why the z is even part not considered. In such a case, the function will be open and its maxima cannot be determined. Please correct me if I am wrong and request you to verify my procedure! Regards, Arpan
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Re: For a particular company, the profit P generated by selling
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09 May 2013, 02:17
\(P = 128 + (\frac{Q2}{4} + 4Q  16)^z\) Can be seen as \(P = 128 + ((\frac{Q8}{2})^2)^z\) \(P=128+(tem\geq{0})^z\) If z is ODD the function has a max in Q=8: we have \(P=128  SecTerm\) => to maximise P we have to minimize the second Term by choosing Q=8 If z is EVEN then the function is a sum: \(P=128 + SecTerm\) => no max value An even value of z will make the second term \(\geq{0}\), so to maximise P we have to increase it as much as we can. I think that the text implied "z is odd and >0", otherwise the question has no solution
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Re: For a particular company, the profit P generated by selling
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09 May 2013, 08:05
emmak wrote: For a particular company, the profit P generated by selling Q units of a certain product is given by the formula P = 128 + (Q^2/4 + 4Q – 16^z), where z > 0. The maximum profit is achieved when Q =
(A) 2 (B) 4 (C) 8 (D) 16 (E) 32 The Question says \(P = 128 + (\frac{Q^2}{4}+4Q16^z)\) I guess if z is to be considered independent of Q in that case we can simply find the maximum value by equating the differential of the equation to Zero.
i.e. \(\frac{d}{dq}\)\((\frac{Q^2}{4}+4Q16^z)\)
=> \(\frac{2Q}{4}+4=0\) as \(16^z\) would be independent of Q
=> Q=8Thanks for the correction. The value of the expression would depend on Z if it were put that way.
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Re: For a particular company, the profit P generated by selling
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10 May 2013, 13:23
The trick is to simplify it out to a certain point, at which it becomes fairly simple as you'll see:
To begin simplifying, the first step is
\(128 + [(\frac{Q}{2})4)*(\frac{Q}{2}+4)]^z\)
Now we take out the 1 on the third part to get two identical parts which are multiplied, which we can combine together by squaring
\(128  ((\frac{Q}{2}4)^2)^z\)
Now it becomes a relatively simple way to find out what the maximum value is for the equation.
we want to find out what Q is when \((\frac{Q}{2}4) = 0\)... this is because we know that the exponent is going to be even (since it's multiplied by 2), so that no matter what it will be possitive. Since we're subtracting from 128 a number that has to be positive, we want the lowest positive number possible in order to have the maximum.
if you solve for \((\frac{Q}{2}4) = 0\) you get that \(Q = 8\)
So choice D.



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Re: For a particular company, the profit P generated by selling
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10 May 2013, 13:32
dave785 wrote: The trick is to simplify it out to a certain point, at which it becomes fairly simple as you'll see:
To begin simplifying, the first step is
\(128 + [(\frac{Q}{2})4)*(\frac{Q}{2}+4)]^z\)
Now we take out the 1 on the third part to get two identical parts which are multiplied, which we can combine together by squaring
\((128  (\frac{Q}{2}4)^2)^z\)
Now it becomes a relatively simple way to find out what the maximum value is for the equation.
we want to find out what Q is when \((\frac{Q}{2}4) = 0\)... this is because we know that the exponent is going to be even (since it's multiplied by 2), so that no matter what it will be possitive. Since we're subtracting a number that has to be positive, we want the lowest positive number possible in order to have the maximum.
if you solve for \((\frac{Q}{2}4) = 0\) you get that \(Q = 8\)
So choice D. The answer depends on z. Example with the answer q=4 that should be wrong \(P = 128 + (\frac{4^2}{4} + 4*4  16)^z\) \(P= 128 + (4)^z\) if z is EVEN example z=2 \(P=128+16=144\), as you see the profit depends on z: if it's even then P is a SUM and has no max value. (if z is ODD your method is correct) Hope it's clear, let me know
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Re: For a particular company, the profit P generated by selling
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10 May 2013, 13:48
dave785 wrote: The trick is to simplify it out to a certain point, at which it becomes fairly simple as you'll see:
we want to find out what Q is when \((\frac{Q}{2}4) = 0\)... this is because we know that the exponent is going to be even (since it's multiplied by 2), so that no matter what it will be possitive. Since we're subtracting from 128 a number that has to be positive, we want the lowest positive number possible in order to have the maximum.
if you solve for \((\frac{Q}{2}4) = 0\) you get that \(Q = 8\)
This is not correct. We don't know whether z is an integer or not. What if z=3/2?
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Re: For a particular company, the profit P generated by selling
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Updated on: 10 May 2013, 14:03
Zarrolou wrote: dave785 wrote: The trick is to simplify it out to a certain point, at which it becomes fairly simple as you'll see:
To begin simplifying, the first step is
\(128 + [(\frac{Q}{2})4)*(\frac{Q}{2}+4)]^z\)
Now we take out the 1 on the third part to get two identical parts which are multiplied, which we can combine together by squaring
\((128  (\frac{Q}{2}4)^2)^z\)
Now it becomes a relatively simple way to find out what the maximum value is for the equation.
we want to find out what Q is when \((\frac{Q}{2}4) = 0\)... this is because we know that the exponent is going to be even (since it's multiplied by 2), so that no matter what it will be possitive. Since we're subtracting a number that has to be positive, we want the lowest positive number possible in order to have the maximum.
if you solve for \((\frac{Q}{2}4) = 0\) you get that \(Q = 8\)
So choice D. The answer depends on z. Example with the answer q=4 that should be wrong \(P = 128 + (\frac{4^2}{4} + 4*4  16)^z\) \(P= 128 + (4)^z\) if z is EVEN example z=2 \(P=128+16=144\), as you see the profit depends on z: if it's even then P is a SUM and has no max value. (if z is ODD your method is correct) Hope it's clear, let me know It doesn't matter whether or not Z is even because we're already squaring it first: \(128  ((\frac{Q}{2}4)^ 2)\) You could then raise that to any value of Z, integer or not, and it would still be positive.
Originally posted by dave785 on 10 May 2013, 13:52.
Last edited by dave785 on 10 May 2013, 14:03, edited 2 times in total.



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Re: For a particular company, the profit P generated by selling
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10 May 2013, 13:57
Quote: \(128  ((\frac{Q}{2}4)^2)^z\)
The very basis of your entire calculation is flawed. When you have taken 1 as the common factor, it has to be\((1)^z\). You have not taken that into consideration.
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Re: For a particular company, the profit P generated by selling
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10 May 2013, 14:04
vinaymimani wrote: Quote: \(128  ((\frac{Q}{2}4)^2)^z\)
The very basis of your entire calculation is flawed. When you have taken 1 as the common factor, it has to be\((1)^z\). You have not taken that into consideration. ah good point. That's what i was doing wrong. Thank you. so it would actually be: \(128 + ((\frac{Q}{2}4)^2)^z\)



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Re: For a particular company, the profit P generated by selling
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10 May 2013, 20:33
dave785 wrote: vinaymimani wrote: Quote: \(128  ((\frac{Q}{2}4)^2)^z\)
The very basis of your entire calculation is flawed. When you have taken 1 as the common factor, it has to be\((1)^z\). You have not taken that into consideration. ah good point. That's what i was doing wrong. Thank you. so it would actually be: \(128 + ((\frac{Q}{2}4)^2)^z\) Preciesly the reason why the value of the complete equation hinges on the value of z. Incase z is odd, the value of (1)^z = 1 and the expression can have a maxima and can be termed as a decreasing function. But incase z is even, the function will be an increasing function since the expression would be of the form CONSTANT + (variable)^2 with no possibile maxima. Regards, Arpan
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Re: For a particular company, the profit P generated by selling
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03 Jul 2013, 10:38
emmak wrote: For a particular company, the profit P generated by selling Q units of a certain product is given by the formula \(P = 128 + (\frac{Q^2}{4} + 4Q  16)^z\), where z > 0. The maximum profit is achieved when Q =
(A) 2 (B) 4 (C) 8 (D) 16 (E) 32 hey here's my approach :: substituting values in place of Q and to keep the whole thing as minimum as possible or we can also do it by equating the q function to 0 and find the roots. so keep 8



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Re: For a particular company, the profit P generated by selling
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03 Jul 2013, 11:47
The trick I mostly use to calculate 'min/max value' or 'min/max value at' is by using derivatives. the Max/Min value occurs when dP __ = 0 dQ \(\frac{dP}{dQ} = z * (  \frac{Q ^ 2}{4} + 4Q  16) ^ (z1) * \frac{d}{dQ} (  \frac{Q^2}{4} + 4Q  16) = 0\) ==> \(( \frac{Q^2}{4} + 4Q  16 ) = 0 or z = 0 or ( \frac{2Q}{4} + 4) =0\) since z > 0 \(( \frac{Q^2}{4} + 4Q  16 ) = 0 or ( \frac{2Q}{4} + 4) =0\) ==> Q = 8 and this is where the Max or Min value occurs.
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Re: For a particular company, the profit P generated by selling
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03 Jul 2013, 17:26
arpanpatnaik wrote: emmak wrote: For a particular company, the profit P generated by selling Q units of a certain product is given by the formula P = 128 + (–Q2/4 + 4Q – 16)^z, where z > 0. The maximum profit is achieved when Q = (A) 2 (B) 4 (C) 8 (D) 16 (E) 32 The answer will be [C], but I believe it will hold for only odd values of z. The above expression can be reduced to P = 128 + {[Q^2 16Q + 64]/2}^z *(1)^z => 128+ [Q8/2]^2z * (1)^z Could someone walk me through ALL the steps in reducing this equation like above? I tried several times to factor it like a quadratic, but I'm way off in my calculations. Thanks in advance, ~ Im2bz2p345



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Re: For a particular company, the profit P generated by selling
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04 Jul 2013, 00:10
Im2bz2p345 wrote: arpanpatnaik wrote: emmak wrote: For a particular company, the profit P generated by selling Q units of a certain product is given by the formula P = 128 + (–Q2/4 + 4Q – 16)^z, where z > 0. The maximum profit is achieved when Q = (A) 2 (B) 4 (C) 8 (D) 16 (E) 32 The answer will be [C], but I believe it will hold for only odd values of z. The above expression can be reduced to P = 128 + {[Q^2 16Q + 64]/2}^z *(1)^z => 128+ [Q8/2]^2z * (1)^z Could someone walk me through ALL the steps in reducing this equation like above? I tried several times to factor it like a quadratic, but I'm way off in my calculations. Thanks in advance, ~ Im2bz2p345 \(P = 128 + (\frac{Q^2}{4} + 4Q  16)^z\). \(P = 128 + (\frac{Q^2+16Q64}{4})^z\) \(P = 128 + (\frac{(Q8)^2}{4})^z\) \(P = 128 + (\frac{(Q8)^2}{2^2})^z\) \(P = 128 + ((\frac{Q8}{2})^2)^z\) \(P = 128 + (1*(\frac{Q8}{2})^2)^z\) \(P = 128 + (1)^z*(\frac{Q8}{2})^{2z}\)
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Re: For a particular company, the profit P generated by selling
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04 Jul 2013, 10:58
Bunuel wrote: \(P = 128 + (\frac{Q^2}{4} + 4Q  16)^z\).
\(P = 128 + (\frac{Q^2+16Q64}{4})^z\)
\(P = 128 + (\frac{(Q8)^2}{4})^z\)
\(P = 128 + (\frac{(Q8)^2}{2^2})^z\)
\(P = 128 + ((\frac{Q8}{2})^2)^z\)
\(P = 128 + (1*(\frac{Q8}{2})^2)^z\)
\(P = 128 + (1)^z*(\frac{Q8}{2})^{2z}\) Thank you SO much Bunuel for taking the time to write this out step by step and clearly!! I understand where I screwed up now. ~ Im2bz2p345



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Re: For a particular company, the profit P generated by selling
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13 Jul 2013, 09:50
Bunuel wrote: Im2bz2p345 wrote: The answer will be [C], but I believe it will hold for only odd values of z. The above expression can be reduced to P = 128 + {[Q^2 16Q + 64]/2}^z *(1)^z => 128+ [Q8/2]^2z * (1)^z Could someone walk me through ALL the steps in reducing this equation like above? I tried several times to factor it like a quadratic, but I'm way off in my calculations. Thanks in advance, ~ Im2bz2p345 \(P = 128 + (\frac{Q^2}{4} + 4Q  16)^z\). \(P = 128 + (\frac{Q^2+16Q64}{4})^z\) \(P = 128 + (\frac{(Q8)^2}{4})^z\) \(P = 128 + (\frac{(Q8)^2}{2^2})^z\) \(P = 128 + ((\frac{Q8}{2})^2)^z\) \(P = 128 + (1*(\frac{Q8}{2})^2)^z\) \(P = 128 + (1)^z*(\frac{Q8}{2})^{2z}\) I understand all the steps, but what is the logic behind the answer being 8 if the entire solution is dependent upon z, and we are not given any indication of what the value of z could be even or odd or integer?? Thanks



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