數學問題 幫忙謝謝?

x+y=30
y=30-x
x^2+y^2
=x^2+(30-x)^2
=2x^2-60x+900
=2(x-15)^2+450
x=y=15 時, 有極小值 450 ....Ans

x^2+y^2
=x^2+(30-x)^2
=x^2+900-60x+x^2
=2x^2-60x+900
=2(x^2-30x)+900
=2(x^2-30x+225-225)+900
=2(x-15)^2-450+900
=2(x-15)^2+450
x^2+y^2的極小值=2(0)+450=450

Let A = x^2 + y^2 = x^2 + (30 - x)^2 = x^2 + 900 + x^2 - 60x. That is A = 2x^2 -60x + 900 = 2(x^2 - 30x + 450) = 2[(x - 15)^2 - 225 + 450] = 2(x - 15)^2 + 450. So max. value of x^2 + y^2 = 450 when x = 15. Correction : should be min. value, not max. value.

用算幾不等式a+b>=2√ab
可以找到xy=225
再用(x+y)^2=900=x^2+y^2+2xy=x^2+y^2+450
所以x^2+y^2極值等於450