Find two positive numbers satisfying the given requirements. the sum of the first and twice the second is 160 and the product is a maximum.

Two positive numbers x, y, such that x + 2y = 160.

The values of x and y that maximize xy.

The product will be maximized when its derivative is zero. We can write the product in terms of a single variable to make solution easier.

... y = (160 -x)/2 . . . . . solve the given relation for y

... f(x) = x·y = x(160 -x)/2 . . . . . substitute the expression for y into the product

... f'(x) = 80 -x = 0 . . . . . . . . . . find the derivative of f(x) and set it to zero

... x = 80 . . . . . . . . . . . . . . . . . add 80

... y = (160 -80)/2 = 40 . . . . . . find the value of y

The first number is 80.

The second number is 40.

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225