Let one number be x. Then, the other number is y = (35-x)

Let P(x)= x² y⁵

Then, we get,

P(x) = 2x(35 –x)⁵ – 5x²(35-x)⁴

= x(35 –x)⁴ [2(35-x)– 5x]

= x(35 –x)⁴ [70-7x]

= 7x(35 –x)⁴(10-x)

Now, P’’(x) = 7(35 –x)⁴ (10-x)+7x[-(35 –x)⁴ – 4(35-x)³ (10-x)]

= 7(35 –x)⁴ (10-x) - 7x(35 –x)⁴ – 28x(35-x)³ (10-x)]

=7(35 –x)³ [(35-x)(10-x) - x(35 –x) – 4x(10-x)]

=7(35 –x)³ [350-45x+x²-35x+x²-40x+4x²]

=7(35 –x)³ [6x²-120x+350]

Now, P’(x) = 0

⇒ x = 0, 35, 10

When x = 0, 35 This will make the product x2y5 equal to 0.

Therefore, x= 0, 35 cannot be possible values of x.

And when x = 10

Then, we have,

P’’(x) =7(35 –10)³[6(10)²-120(10)+350]

= 7(25)³ [-250]<0

Then, by second derivative test,

x = 10 and y = 35 -10 = 25 is the point of local maxima of P.

Therefore, the required number are 10 and 25.