Question

Two numbers have a difference of 20. How do you find the numbers if the sum of their squares is a minimum?

Answer 1

Here we assume two numbers as two variables and use the concept that numbers have a difference of 20 to form the first equation. Then form up an equation of sum of squares of these two numbers and substitute the value of one variable in terms of another using the first equation. We can calculate the minimum value using the first derivative of equation, i.e. by equating it to 0.

• Differentiation of ${x^n}$ with respect to x is given by $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
• If the second derivative of a function is less than 0 then it has a local maxima and if it is greater than 0 then it has local minima.

Complete step-by-step answer:
Let us take two numbers as ‘x’ and ‘y’.
We are given that numbers have difference 20, this means we subtract one number from another we get the value 20
$\Rightarrow x - y = 20$
Shift one variable to right hand side of the equation to form one variable in terms of another
$\Rightarrow x = 20 + y$ … (1)
Now we know the sum of squares of two numbers is minimum
$\Rightarrow {x^2} + {y^2}$ is minimum
Use the substitution from equation (1) to open the value in left hand side of the equation
$\Rightarrow {x^2} + {y^2} = {(20 - y)^2} + {y^2}$
Open right hand side of the equation using the identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$
Shift all the constant values to one side
$\Rightarrow {x^2} + {y^2} = {(20)^2} + {(y)^2} - 2 \times 20 \times y + {y^2}$
$\Rightarrow {x^2} + {y^2} = 400 + {y^2} - 40y + {y^2}$
Add like terms
$\Rightarrow {x^2} + {y^2} = 2{y^2} - 40y + 400$ … (2)
Now we are given that sum of squares of two numbers is minimum
i.e. the right hand side of equation (2) should be minimum.
We will calculate differentiation of the equation with respect to y and equate it to zero to find the minimum value.
$\Rightarrow \dfrac{d}{{dy}}\left( {2{y^2} - 40y + 400} \right) = 4y - 40$
Equating derivative to 0
$\Rightarrow 4y - 40 = 0$
Shift constant value to right hand side of the equation
$\Rightarrow 4y = 40$
Cancel same factors from both sides of the equation
$\Rightarrow y = 10$
Substitute the value of y in equation (1)
$\Rightarrow x = 20 + 10$
$\Rightarrow x = 30$

$\therefore$The two numbers are 10 and 30.

Note:
Students many times make the mistake of not changing the sign when shifting the values from one side of the equation to another side, always keep in mind the sign changes from positive to negative and vice versa when we shift a number from one side to another side of the equation.