Question $3$ $\newline$ Given that $-6 \leq x \leq 3$ and $1 \leq y \leq 5$ , find the greatest possible value of $x^{2}+y^{2}$ .

Full solution

Q. Question

3

\newline

Given that

-6 \leq x \leq 3

and

1 \leq y \leq 5

, find the greatest possible value of

x^{2}+y^{2}

Given Ranges: We are given the ranges for $x$ and $y$ : $-6 \leq x \leq 3$ $1 \leq y \leq 5$ We need to find the greatest possible value of the expression $x^{2}+y^{2}$ .To maximize the expression, we need to find the maximum values of $x^{2}$ and $y^{2}$ within the given ranges.
Maximize $x^{2}$ : For $x^{2}$ to be maximum, we need to find the value of $x$ that gives the largest square within the range $-6 \leq x \leq 3$ . Squaring a negative number or a positive number both give a positive result, and the square of a larger absolute value is greater. Therefore, the maximum value of $x^{2}$ occurs at $x = -6$ . Calculate $x^{2}$ when $x = -6$ . $(-6)^{2} = 36$
Maximize $y^2$ : For $y^2$ to be maximum, we need to find the value of $y$ that gives the largest square within the range $1 \leq y \leq 5$ . Since squaring a number gives a positive result and the function $y^2$ is increasing for positive $y$ , the maximum value of $y^2$ occurs at $y = 5$ . Calculate $y^2$ when $y = 5$ . $y^2$ $0$
Calculate Maximum Value: Now, we add the maximum values of $x^{2}$ and $y^{2}$ to find the greatest possible value of the expression $x^{2}+y^{2}$ . $\newline$ $36$ (from $x^{2}$ ) + $25$ (from $y^{2}$ ) = $61$