Respuesta :
Answer:
The least possible value is -2, obtained for a = -1 and b = 1.
Step-by-step explanation:
We want the minimum of [tex] \displaystyle\left(\frac{1}{a} \frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) [/tex] , where
- [tex] -5 \leq a \leq -1 [/tex]
- [tex] 1 \leq b \leq 3 [/tex]
First, lets simplify the expression given, we use common denominator on the second part, using ab as the common denominator. We obtain
[tex] \frac{1}{b} - \frac{1}{a} = \frac{a-b}{ab} [/tex]
As a result
[tex] \displaystyle\left(\frac{1}{a} \frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) = \frac{a-b}{(ab)^2} [/tex]
we need the minimum of the function
[tex] f(a,b) = \frac{a-b}{(ab)^2} [/tex]
with the restrictions [tex] -5 \leq a \ -1, 1 \leq b \leq 3 [/tex]
First, we calculate the gradient of f and find where it takes the zero value.
[tex]\nabla{f} = (f_a,f_b) [/tex]
with
[tex]f_a = \frac{(ab)^2 - (a-b) 2ab^2}{(ab)^2} = -1 + 2 \frac{b}{a}[/tex]
Since it has the reversed sign, we get
[tex]f_b = - (-1 + 2 \frac{a}{b}) =1 - 2 \frac{a}{b}[/tex]
In order for [tex] \nabla{f} [/tex] to be zero, we need both [tex] f_a [/tex] and [tex] f_b [/tex] to be zero, observe that
[tex]f_b = 0 \, \rightarrow 1 -2\frac{a}{b} = 0 \, \rightarrow 1 = 2 \frac{a}{b} \, \rightarrow b = 2a [/tex]
Which is impossible with the given restrictions. Hence, the minimum is realized in the border.
If we fix a value a₀ for a, with a₀ between -5 and -1 the function g(b) = f(a₀,b) wont have a minimum for b in [1,3] because the partial derivate of f over b didnt reach the value 0 in the restrictions given. On the other hand. by making a similar computation that before, we can obtain that the partial derivate of f over the variable a doesnt reach the value 0 either. This means that f doesnt reach the minimum on the sides. As a consecuence, it reach a minimum on the corners.
The 4 possible corner values are (-5,1), (-5,3), (-1,1) and (-1,3)
[tex] f(-5,1) = \frac{-6}{25} = -0.24 [/tex]
[tex] f(-5,3) = \frac{-8}{225} = -0.0355 [/tex]
[tex] f(-1,1) = \frac{-2}{1} = -2 [/tex]
[tex] f(-1,3) = \frac{-4}{9} = -0.444 [/tex]
Clearly the least possible value between the four corners is -2.
