Respuesta :

let's say the numbers are "a" and "b"

thus [tex]\bf y=ab\qquad \begin{cases} a+b=80\\ b=80-a \end{cases}\implies y=a(80-a)\implies y=80a-a^2\\\\ -----------------------------\\\\ \cfrac{dy}{da}=80-2a[/tex]

set the derivative to 0, and check the critical points, there's only one anyway

and do a first-derivative test, to see if it's a maximum
xy=max
x+y=80
minus x both sides to solve for y
y=80-x
subsitute

(x)(80-x)=max
80x-x²=max
take the derivitive
80-2x=max'
find where it equalt 0
80-2x=0
80=2x
40=x
at x=40
then y=80-40=40

the numbers are 40 and 40


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