4.7 Applied optimization problems  (Page 4/8)

Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let $L$ be the length of the rectangle and $W$ be its width. Let $A$ be the area of the rectangle.

Step 2: The problem is to maximize $A.$

Step 3: The area of the rectangle is $A=LW.$

Step 4: Let $\left(x,y\right)$ be the corner of the rectangle that lies in the first quadrant, as shown in [link] . We can write length $L=2x$ and width $W=2y.$ Since $\frac{x^2}{4+y^2=1}$ and $y>0,$ we have $y=\sqrt{\frac{1-x^2}{4}}.$ Therefore, the area is

Step 5: From [link] , we see that to inscribe a rectangle in the ellipse, the $x$ -coordinate of the corner in the first quadrant must satisfy $0<x<2.$ Therefore, the problem reduces to looking for the maximum value of $A\left(x\right)$ over the open interval $\left(0,2\right).$ Since $A\left(x\right)$ will have an absolute maximum (and absolute minimum) over the closed interval $\left[0,2\right],$ we consider $A\left(x\right)=2x\sqrt{4-x^2}$ over the interval $\left[0,2\right].$ If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.

Step 6: As mentioned earlier, $A\left(x\right)$ is a continuous function over the closed, bounded interval $\left[0,2\right].$ Therefore, it has an absolute maximum (and absolute minimum). At the endpoints $x=0$ and $x=2,$ $A\left(x\right)=0.$ For $0<x<2,$ $A\left(x\right)>0.$ Therefore, the maximum must occur at a critical point. Taking the derivative of $A\left(x\right),$ we obtain

To find critical points, we need to find where $A\prime \left(x\right)=0.$ We can see that if $x$ is a solution of

then $x$ must satisfy

Therefore, $x^2=2.$ Thus, $x=\text{±}\sqrt{2}$ are the possible solutions of [link] . Since we are considering $x$ over the interval $\left[0,2\right],$ $x=\sqrt{2}$ is a possibility for a critical point, but $x=\text{−}\sqrt{2}$ is not. Therefore, we check whether $\sqrt{2}$ is a solution of [link] . Since $x=\sqrt{2}$ is a solution of [link] , we conclude that $\sqrt{2}$ is the only critical point of $A\left(x\right)$ in the interval $\left[0,2\right].$ Therefore, $A\left(x\right)$ must have an absolute maximum at the critical point $x=\sqrt{2}.$ To determine the dimensions of the rectangle, we need to find the length $L$ and the width $W.$ If $x=\sqrt{2}$ then

Therefore, the dimensions of the rectangle are $L=2x=2\sqrt{2}$ and $W=2y=\frac{2}{\sqrt{2}}=\sqrt{2}.$ The area of this rectangle is $A=LW=\left(2\sqrt{2}\right)\left(\sqrt{2}\right)=4.$