Question 964813
Find the minimum value of Z where Z is defined as,
{{{Z=xy}}}
.
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From the equation,
{{{2y=42-3x}}}
{{{y=21-(3/2)x}}}
Substituting,
{{{Z=x(21-(3/2)x)}}}
{{{Z=21x-(3/2)x^2}}}
Convert Z to vertex form, then you can find the extremum.
{{{Z=-(3/2)x^2+21x}}}
{{{Z=-(3/2)(x^2-14x)}}}
{{{Z=-(3/2)(x^2-14x+49)+(3/2)(49)}}}
{{{Z=-(3/2)(x-7)^2+147/2}}}
The value at the vertex is a maximum because of the {{{-3/2}}} multiplier.
It occurs when {{{x=7}}} and the maximum value is {{{147/2}}}.
There is no minimum for the product, only a maximum.
Check your problem setup and repost if needed.
*[illustration maxc.JPG].