Find the curve of the maximum value of work done?

Question

Suppose $C$ is a simple close curve (i.e. it doesn’t intersect itself) in the first quadrant. If $F = (y^2/2 + x^2y, -x^2 + 8x)$, find the curve that produces the maximum amount of work done by $F$. What is the maximum value of work?

Answer 1

The maximization problem can be reduced to the case of a region in the first quadrant of $\mathbb{R}^2$ by using Green's Theorem: $$ \oint_C F\cdot ds=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$$ So, we can maximize the integral over $D$ of $$ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=-2x+8-(y+x^2).$$ This is achieved precisely by integrating over the entirety of the region $D$ on which the integrand above is non-negative: $$D=\{(x,y)\in \mathbb{R}^2: 8-2x-y-x^2\ge 0,\text{and}\:x,y\ge 0\}.$$ So, we need to figure out which points in the first quadrant have $y\le 8-2x-x^2$. Completing the square yields $y-9\le -1-2x-x^2=-(x^2+2x+1)=-(x+1)^2.$ So, $$ D=\{(x,y)\in \mathbb{R}^2: y\le -(x-1)^2+9\}.$$ We see that $D$ is bounded by a parabola opening in the negative $y$ direction and the coordinate axes. So, the boundary curve $C=\partial D$ is given by parametrizing the union $$\underbrace{\{(0,t):0\le t\le8\}}_{\text{the portion of the $y-$axis}}\cup\underbrace{\{(t,0): 0\le t\le 4\}}_{\text{the portion of the $x-$axis}}\cup\underbrace{\{(x,y):y=-(x-1)^2+9,\text{and}\:0\le x\le 4\}}_{\text{the portion of the curve}}.$$ At this point, you should compute $$ \oint_{\partial D} F\cdot ds=\iint_{D}(-2x+8-y-x^2)dA$$ as this maximizes the work (i.e. the line integral on the left).