Show that y reaches its largest feasible value when y=x/m.

01 / Standard#Your turn

x,y>0,\quad0<m<1,\quad y^2-2mxy+x^2=a^2

Official paper · jm01-2021 · II.3(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Divide the identity in part (a) by the positive number 1−m².

Hint 2

The subtractive square vanishes precisely when x=my.

Worked solution

Obtain a sharp upper bound for positive y.
$y^2=\frac{a^2}{1-m^2}-\frac{(x-my)^2}{1-m^2}\le\frac{a^2}{1-m^2}$
Equality is attained at positive coordinates because 0<m<1 and feasibility forces a≠0.
$y=\frac{|a|}{\sqrt{1-m^2}},\quad x=my>0$
Since m is nonzero, the equality relation is equivalent to the printed form.
$x=my\iff y=x/m$

The maximum occurs when x=my, equivalently y=x/m.

Checks and common pitfalls: The fraction x/m must not be misread as the product mx.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Obtain a sharp upper bound for positive y.
$y^2=\frac{a^2}{1-m^2}-\frac{(x-my)^2}{1-m^2}\le\frac{a^2}{1-m^2}$
Equality is attained at positive coordinates because 0<m<1 and feasibility forces a≠0.
$y=\frac{|a|}{\sqrt{1-m^2}},\quad x=my>0$
Since m is nonzero, the equality relation is equivalent to the printed form.
$x=my\iff y=x/m$

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

2021 JM01