Find the minimum of t=x²+y² over the feasible region.

01 / Standard#Your turn

3x+2y\ge13,\quad x\le5,\quad2x-2y+3\ge0

Official paper · jm01-2023 · II.5(c) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use the constraint 3x+2y≥13.

Hint 2

Apply Cauchy–Schwarz and check its equality point.

Worked solution

Bound the squared distance.
$13(x^2+y^2)\ge(3x+2y)^2\ge169\implies t\ge13$
Equality requires (x,y) proportional to (3,2), with 3x+2y=13; the resulting point is feasible.
$(x,y)=(3,2),\quad x\le5,\quad2x-2y+3=5\ge0$

Minimum 13 at (3,2).

Checks and common pitfalls: The nearest point may lie on an edge rather than at a vertex.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Bound the squared distance.
$13(x^2+y^2)\ge(3x+2y)^2\ge169\implies t\ge13$
Equality requires (x,y) proportional to (3,2), with 3x+2y=13; the resulting point is feasible.
$(x,y)=(3,2),\quad x\le5,\quad2x-2y+3=5\ge0$

Think first. Reveal a hint when the class is ready.

2023 JM01