How do you find two positive numbers whose sum is 300 and whose product is a maximum?
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To find two positive numbers whose sum is 300 and whose product is a maximum, you can use the method of calculus called optimization. Let the two numbers be ( x ) and ( y ), then we have the following constraints:
- ( x + y = 300 ) (sum constraint)
- ( P = xy ) (product)
To maximize the product ( P ), we need to find the critical points of the product function ( P = xy ) subject to the constraint ( x + y = 300 ). We can use the method of Lagrange multipliers to solve this constrained optimization problem.
The Lagrangian function is given by:
[ L(x, y, \lambda) = xy - \lambda(x + y - 300) ]
Taking partial derivatives with respect to ( x ), ( y ), and ( \lambda ) and setting them equal to zero, we obtain:
[ \frac{\partial L}{\partial x} = y - \lambda = 0 ] [ \frac{\partial L}{\partial y} = x - \lambda = 0 ] [ \frac{\partial L}{\partial \lambda} = -(x + y - 300) = 0 ]
Solving these equations simultaneously will give us the values of ( x ), ( y ), and ( \lambda ). Substituting these values back into the original constraints will give us the maximum product.
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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.
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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