Question

For any real numbers a, b and c, find the smallest value of the expression $3a^2+27b^2+5c^2-18ab-30c+237.$3a2+27b2+5c2−18ab−30c+237

Answer 1

The given expression:

$3a^2+27b^2+5c^2-18ab-30c+237$
Rearranging the terms we get,
$3a^2+27b^2-18ab+5c^2-30c+237$

Taking $3$ common from the coefficients of $a^2,b^2$ and $ab$ terms. Also taking $5$ common from the coefficients of $c^2$ and $c$ terms. The expression then becomes:

$3(a^2+(3b{)}^2–2\cdot a\cdot 3b)+5(c^2–2\cdot c\cdot 3)+237$

The term in the first braces can be condensed into $(a-3b{)}^2$

The term in the second braces needs a $(3^2=9)$ term to be condensed into $(c-3{)}^2$

So we add $9$ inside the second braces and subtract $5\cdot 9$ from $237$ as in order to get $9$ inside the second braces terms we need to enter it as a product with $5$
Hence, term to be introduced in the second braces $=5\cdot 9=45$, subsequently, $45$ would also get reduced from $237,$ in order to balance the equation

$3(a-3b{)}^2+5(c-3{)}^2+237–45$

The final expression becomes:

$3(a-3b{)}^2+5(c-3{)}^2+192$

The lowest value of this expression occurs when each one of its terms attains its lowest, as the terms are independent of each other.

This happens when first two terms become zero (as the square of any real number cannot be negative) i.e.

$a-3b=0$ and $c-3=0$ , which gives us the lowest value of the given expression to be $192$ when $a=3b$ and $c=3$