# Optimization: sum of squares | Applications of derivatives | AP Calculus AB | Khan Academy | Summary and Q&A

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January 31, 2013
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Optimization: sum of squares | Applications of derivatives | AP Calculus AB | Khan Academy

## TL;DR

The smallest sum of squares of two numbers with a negative product of -16 is 32.

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### Q: How can we find the smallest sum of squares of two numbers with a negative product of -16?

We can express the sum of squares in terms of x and y, and given their product is -16, we can substitute y with -16/x. By minimizing this expression, we find the smallest sum of squares.

### Q: What is the derivative of the sum of squares function?

The derivative of the sum of squares with respect to x is 2x*(-2)2x + 256(-2). Simplifying, we get -512x^(-3).

### Q: What are the critical points of the sum of squares function?

The only critical point is x=4, since x=0 would result in an undefined y. This critical point minimizes the sum of squares.

### Q: How can we confirm that x=4 results in a minimum value?

We can use the second derivative test. Taking the second derivative of the sum of squares function, we find that it is always positive, indicating a concave upwards shape and confirming the minimum point at x=4.

## Summary & Key Takeaways

• The goal is to find the smallest sum of squares of two numbers whose product is negative 16.

• By expressing the sum of squares as a function of x and y, and using the given information about their product, we can rewrite the expression in terms of x only.

• Taking the derivative of the function and finding the critical points, we find that the only critical point is x=4, which minimizes the sum of squares.

• The minimum sum of squares is 32, with x=4 and y=-4.