Minimize Shaded Area of Composite Rectangular Shapes

Question

For which length x is the shaded area minimized? a) Rectangle with shaded inner rectangle, outer dimensions 5x3, inner rectangle offset by x on each side; b) Shaded parallelogram in 5x3 rectangle, offset by x; c) Shaded area in 5x3 rectangle with two triangles cut out at top, base x

Accepted Answer

a) x=1.25, minimum area=7.5; b) x=0, minimum area=0; c) x=5, minimum area=7.5

Answer

Step 1 (part a): Shaded area $A(x)=(5-2x)(3-2x)=4x²-16x+15$. Find vertex of quadratic: $x=-\frac{b}{2a}=\frac{16}{8}=2$, but wait, domain 0<x<1.5. Wait correct area: outer area=15, shaded area=15 - area of unshaded? No, shaded is inner rectangle: $A(x)=(5-2x)(3-2x)$. Vertex at x=2 is outside domain, so minimum at x=0 (area=15) or x=1.5 (area=0). Wait, no, the diagram shows shaded as inner rectangle with x offset, so correct area is $(5-2x)(3-2x)$, minimum at x approaching 1.5, area approaching 0. Wait, re-interpret: shaded area is the border, so $A(x)=15-(5-2x)(3-2x)=-4x²+16x$. Vertex at x=2, but domain 0<x<1.5, so maximum at x=1.5, minimum at x=0 (area=0) Step 1 (part b): Shaded parallelogram area $A(x)=x\times3=3x$. Minimum at x=0, area=0 Step 1 (part c): Shaded area $A(x)=15 - 2\times\frac{1}{2}x\times3=15-3x$. Minimum at x=5 (max x), area=15-15=0