Answer:
The least possible value is -2, obtained for a = -1 and b = 1.
Explanation:
We want the minimum of
, where
First, lets simplify the expression given, we use common denominator on the second part, using ab as the common denominator. We obtain
As a result
![\displaystyle\left((1)/(a) (1)/(b)\right)\left((1)/(b)-(1)/(a)\right) = (a-b)/((ab)^2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/h6qyhciswxli98lurhlypdw2lskm9ie7z8.png)
we need the minimum of the function
with the restrictions
![-5 \leq a \ -1, 1 \leq b \leq 3](https://img.qammunity.org/2021/formulas/mathematics/high-school/y63pgv4l6115levkul7s7ldqhhgdhssuef.png)
First, we calculate the gradient of f and find where it takes the zero value.
![\\abla{f} = (f_a,f_b)](https://img.qammunity.org/2021/formulas/mathematics/high-school/5zmb38merlz9p7dmmj2zuev0qd2rav21k6.png)
with
![f_a = ((ab)^2 - (a-b) 2ab^2)/((ab)^2) = -1 + 2 (b)/(a)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rrppdiywe9n2pg5klb18fqny4n6b63ld0z.png)
Since it has the reversed sign, we get
![f_b = - (-1 + 2 (a)/(b)) =1 - 2 (a)/(b)](https://img.qammunity.org/2021/formulas/mathematics/high-school/e85uobp9ei1delvwo6hahoym707h1gd0v5.png)
In order for
to be zero, we need both
and
to be zero, observe that
![f_b = 0 \, \rightarrow 1 -2(a)/(b) = 0 \, \rightarrow 1 = 2 (a)/(b) \, \rightarrow b = 2a](https://img.qammunity.org/2021/formulas/mathematics/high-school/40v4axg0doyepam0kn43c4x4z3ae5aai4u.png)
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)
![f(-5,1) = (-6)/(25) = -0.24](https://img.qammunity.org/2021/formulas/mathematics/high-school/w9ady0andfx4o9ryi5idji9eayloc64haw.png)
![f(-5,3) = (-8)/(225) = -0.0355](https://img.qammunity.org/2021/formulas/mathematics/high-school/lfeq9kj7h56t657mkb744ed9mbrj8b4bkv.png)
![f(-1,1) = (-2)/(1) = -2](https://img.qammunity.org/2021/formulas/mathematics/high-school/47wq0gsi2ryzgbyrb03rded6i69ajc54bh.png)
![f(-1,3) = (-4)/(9) = -0.444](https://img.qammunity.org/2021/formulas/mathematics/high-school/7qy0bgqt57xrj5ey3jsu46ugr7ynm142ei.png)
Clearly the least possible value between the four corners is -2.