calculus optimization easy?

Question

Find the point or points on the graph of y = sqrt(3x + 40) with
-10<=x<=10 that is/are closest to the origin.

Answer 1

When a radical appears in an equation, such as this, it means the positive square root. So we would only expect one solution.
Let (x, y) be any point on the graph, then the distance from the point to the origin is:
D = √(x^2 + y^2)
y = √(3x + 40)
y^2 = 3x + 40
Plug in for y^2:
D = (x^2 + 3x + 40)
Take the derivative and set it equal to zero:
D ' = (1/2)(x^2 + 3x + 40)^(-1/2)(2x + 3) = 0
2x + 3 = 0
x = -3/2
y = √(3(-3/2) + 40)
y =√ (-9/2 + 40)
y = √(71/2)
One point (-3/2, √(71/2))

Answer 2

enable x and y the size (eq1) A=xy=125000 ,or x=125000 /y; (eq2)P=2x+2y,or P=2(125000/y)+2y then get dP/dy=0,which consists of y because of the fact the only unknown.then scientific care y.given this y, u can examine x. in case you are trying this, you will detect that the smallest perimeter is closest to an oblong. consequently, x=y=353.55 m