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emmak
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+ [Q-8/2]^2z * (-1)^z

for all values of z, [Q-8/2]^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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\(P = 128 + (-\frac{Q2}{4} + 4Q - 16)^z\)

Can be seen as

\(P = 128 + (-(\frac{Q-8}{2})^2)^z\)
\(P=128+(-tem\geq{0})^z\)

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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arpanpatnaik
emmak
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+ [Q-8/2]^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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dave785
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.
­

How are you assuming 
Q/2 + 4 = 0????
 
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Apeksha2101

dave785
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 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.
­

How are you assuming 
Q/2 + 4 = 0????


 

We want to maximize ­\(128 - ((\frac{Q}{2}-4)^2)^z\). Since the square of a number is always positive or 0, then the value of \(128 - ((\frac{Q}{2}-4)^2)^z\) will be maximized when you subtract the least value from 128, so when you subtract 0.­
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Alternate method may be if you are running out of time.

From the equation its pretty clear that we should maximize
f(Q) = -Q^2/4 + 4Q -16

If you know how to calculate derivative(for equations of smaller order its pretty simple)
provided its an continuous uphill and down.
f'(Q)=-2Q/4 + 4
then f'(Q) = 0
Q=8­
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But z could be even

since z > 0 , if z is even , Max P = 128 + |(−Q^2)/4+4Q−16|^z ??
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Why isnno 9nw just inserting numbers start from c
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