# Find two positive numbers that satisfy the given requirements. The sum of the first number squared and the second number is 60 and the product is a maximum?

The numbers are

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Let the two positive numbers be ( x ) and ( y ). To find the maximum product, we need to maximize the product ( P = xy ).

Given that ( x^2 + y = 60 ), we can solve for ( y ) to get ( y = 60 - x^2 ). Substituting this into the expression for ( P ), we get ( P = x(60 - x^2) = 60x - x^3 ).

To find the maximum of ( P ), we take its derivative with respect to ( x ), set it equal to zero, and solve for ( x ): [ \frac{dP}{dx} = 60 - 3x^2 = 0 ] [ 3x^2 = 60 ] [ x^2 = 20 ] [ x = \sqrt{20} ] [ x = 2\sqrt{5} ]

Substitute ( x = 2\sqrt{5} ) into the equation for ( y ) to find the corresponding value: [ y = 60 - (2\sqrt{5})^2 = 60 - 20 = 40 ]

So, the two positive numbers that satisfy the given requirements are ( x = 2\sqrt{5} ) and ( y = 40 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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