Question

Find the absolute maximum and minimum values of\(f\)the set\(D\).

Short answer

Absolute minimum is\(0\)& absolute maximum is\(13\) .

Step-by-step solution

Definition

The highest point of a function within the entire domain is known as the absolute maxima of the function whereas the lowest point of the function within the entire domain of the function, is known as the absolute minima of the function.

Region

The region is shown below:

Find critical point

Consider the function \(f(x,y) = 4x + 6y - {x^2} - {y^2}\)

The partial derivative is:

\(\begin{aligned}{l}{f_x} = 4 - 2x\\{f_y} = 6 - 2y\end{aligned}\)

Put\({f_x} = 0,\;\;\;{f_y} = 0\)

\(\begin{aligned}{l}2(2 - x) = 0\\2(3 - y) = 0\end{aligned}\)

On solving these equations we find the critical points \((2,3)\)

\(f(2,3) = 14\)

Function on boundary

On boundary line\({L_1},y = 0\)and

\(f(x,0) = 4x - {x^2},\;\;\;0 \le x \le 4\)

The maximum value is\(f(0,0) = f(4,0) = 0\)value\(f(2,0) = 4\)and minimum value

On boundary line \({L_2},x = 4\)and

\(\begin{aligned}{c}f(4,y) = 16 + 6y - 16 - {y^2}\\ = - {y^2} + 6y\end{aligned}\)

The maximum value is\(f(4,3) = 9\)and minimum value\(f(4,0) = 0\).

Function on boundary

On boundary line\({L_3},y = 5\) and

\(\begin{aligned}{c}f(x,5) = 4x + 30 - {x^2} - 25\\ = - {x^2} + 4x + 5\end{aligned}\)

The maximum value \(f(2,5) = 9\)and minimum value\(f(0,5) = f(4,5) = 5\)

On boundary line\({L_4},x = 0\)

\(f(0,y) = 6y - {y^2},0 \le y \le 5\)

The maximum value\(f(0,3) = 9\)and minimum value\(f(0,0) = 0\).

So, the absolute maximum is\(13\)and absolute minimum is\(\;0\).