The largest possible value of p = 130
What is Integer?
Zero, a positive natural number, or a negative integer denoted by a minus sign are all examples of integers. The inverse additives of the equivalent positive numbers are the negative numbers. The boldface Z is a common way to represent the set of integers in mathematical terms.
Given,
2mnp = (m + 2) (n+2)(p+2)
lets first solve p
⇒(2mn)p = p((m + 2) (n+2) + 2 (m+2)(n+2)
⇒[2mn-(m+2)(n+2)p
=2(m+2)(n+2)
⇒p = [tex]\frac{2 (m+2)(n+2)}{mn-2n-2m-4}[/tex]
⇒p = [tex]\frac{2(m+2)(n+2)}{(m-2)(n-2)-8}[/tex]
Since it is obvious that we wish to reduce the denominator, we test (m-2)(n-2) - 8 = 1.
(m-2)(n-2) = 9. The pairs of 9 that can exist are (1, 9) (3,3). The results are, correspondingly, m = 3, n = 11, and m = 5, n = 5.
When we substitute the first pair into the numerator, we get 130, whereas the second pair gets 98. We now verify that 130 is the best value, setting a=m-2 and b=n-2 to facilitate calculations.
Since, 0 ≤(a-1)(b-1) ⇒ a+b≤ab+1
we have,
p = [tex]\frac{2(a+4)(b+4)}{ab-8}[/tex]
= [tex]\frac{2ab+8(a+b++32}{ab-8} \leq \frac{2ab+8(ab+1)+32}{ab-8}[/tex]
= 10 +[tex]\frac{120}{ab-8}[/tex]≤130
Where we see (m,n)=(3,11) gives us our maximum value of 130
Remember that 0 ≤ (a-1)(b-1) assumes m,n ≥3, but there is clear as [tex]\frac{2m}{m+2}[/tex] = [tex]\frac{(n+2)(p+2)}{np}[/tex] > 1 and similarly for n
we state the denominator differently as we solve for p.
p = [tex]\frac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)}[/tex] ⇒[tex]\frac{1}{p}[/tex] = [tex]\frac{1}{2}[/tex] - [tex]\frac{2(m+n+2)}{(m+2)(n+2)}[/tex]
Here it suffices to maximise [tex]\frac{m+n+2}{(m+2)(n+2)}[/tex] under the conditions that p is a positive integer.
Then, m+n+2/(m+2)(n+2) > 1/2 for m = 1,2, we fix m =3
⇒ 1/p = 1/2-[tex]\frac{2(n+2)}{5(n+2)}[/tex]
= n-10/10(n+2)
where we let n=11 to achieve p = 130
The largest possible value is 130
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