Question 346239
1.{{{b+h=16}}}
2.{{{A=(1/2)bh}}}
From eq. 1,
{{{b=16-h}}}
Substitute into eq. 2,
{{{A=(1/2)(16-h)h}}}
{{{A=(1/2)(16h-h^2)}}}
{{{A=-(1/2)h^2+8h}}}
Convert to vertex form to find the maximum of A, {{{A=a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex.
{{{A=-(1/2)h^2+8h}}}
{{{A=-(1/2)(h^2-16h)}}}
{{{A=-(1/2)(h^2-16h+64)+64/2}}}
{{{A=-(1/2)(h-8)^2+32}}}
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{{{drawing(300,300,-2,18,-10,40,grid(1),blue(line(8,-200,8,200)),circle(8,32,0.5),graph(300,300,-2,18,-10,40,-(1/2)(x-8)^2+32))}}}
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The vertex is ({{{8}}},{{{32}}}).
The coefficient of the {{{h^2}}} term is negative so the value at the vertex is a maximum.
{{{Amax=32}}}cm^2
{{{highlight(h=8)}}}cm
From eq. 1,
{{{b+8=16}}}
{{{highlight(b=8)}}}cm