Maximize Area of Rectangle Formed by Function $f(x)=0.5x^2 -4x +7.5$ and Axes

Question

Given function $f(x)=0.5x^2 -4x +7.5$, defined for $0 ≤ x ≤3$. A point P on the graph forms a rectangle with the axes and lines parallel to the axes. a) Sketch the situation; b) For which x is the area of the rectangle maximum?

Accepted Answer

b) $x=1$; maximum area = 4

Answer

Step 1: Define the rectangle's width as $x$ and height as $f(x)$, so area function $A(x)=x\cdot f(x)=x(0.5x^2-4x+7.5)=0.5x^3-4x^2+7.5x$ Step 2: Find first derivative $A'(x)=1.5x^2-8x+7.5$, set to 0: $1.5x^2-8x+7.5=0$ → multiply by 2: $3x^2-16x+15=0$ Step 3: Solve quadratic equation: $x=\frac{16\pm\sqrt{256-180}}{6}=\frac{16\pm\sqrt{76}}{6}=\frac{16\pm2\sqrt{19}}{6}=\frac{8\pm\sqrt{19}}{3}$. Only $x=\frac{8-\sqrt{19}}{3}≈1$ is in $0≤x≤3$ Step 4: Verify it's a maximum using second derivative $A''(x)=3x-8$, $A''(1)=-5<0$ (local maximum). Calculate $A(1)=0.5-4+7.5=4$