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Help please

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Is this calculus or no? There are multiple ways to approach and justify this and the best way depends on what tools you're allowed to use.
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You'll need to use calculus for this

P = xy

P = (8-2y)y = 8y - 2y^2

Differentiate now

P' = 8 - 4y

Now to find if this is the maxima or minima, you'll need to get the second order derivative, and if it's negative then it's maxima, if positive then minima

P'' = -4 which is negative. So the value you'll get is the maximum value, not the minimum

Now equate P' to 0 to get the critical points

P' = 8-4y = 0

y = 2

x = 8-2y = 4

These are the real numbers. And the product will be 8, the maximum product
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>Determine the real numbers x and y satisfying x + 2y = 8 and such that the product P(x, y) = xy is maximized. Give justification that your solution is actually a maximum.


x=8-2y

xy=y(8-2y)=8y-2y^2

This is a parabola that opens down and has a max. at the vertex: -b/(2a)

f(y)=-2y^2 +8y

vertex: y=-8/(2(-2))=2

Now find x

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