Probem17+Sabrina

Use the word "maximize" or "minimize" prior to the objective function! By the way, there is a trick with square root functions. since we know that for y = √x, as x increases y also increases, we can simply find the maximum value of the radicand (inside) , and then assume that that will give us the maximum value of the square root.

Also consider the domain of x...what is the distance from the origin as x gets close to 0? (I don't remember --- must x be positive?) what happens as x approaches infinity? In other words, look at the end behavior for your distance function.

The objective of this problem is to find the point on the line closest to the origin. - in this problem you mus use the distance formula - point one = (0,0) - point two = (x,y)

D= √((x2²-x1²) +(y2²-y1²))

Plug in points

D= √(x² +y²)

We also know that y = 4x+7

D= √(x² +(4x+7)²) D= √(x² +16x^2 + 56x + 49) D= √(17x^2 + 56x + 49)

Now take the derivative and set that equal to 0

D' = 1/2(17x^2 + 56x + 49)^-1/2 (34x+56) D'= 34x+56 / 2√(17x^2 + 56x + 49) 34x + 56 = 0 x = -1.65

plug x back into equation y = 4x+7 y = .412

(im not sure how to check this answer)

 So we're saying that the Point (-1.65, .412) is the closest the fuction gets to the origin. Perhaps a graph could be some sort of visual check of our work?



The red line is the original fuction. The blue line is a line the passes through the origin and what we found to be the closest point on the function to the origin. Visually, (-1.65, .412) looks like the closest point on the line y=4x+7 to the origin.

lim √(17x^2 + 56x + 49) = infinity x->infinity

lim √(17x^2 + 56x + 49) = infinity x-> -infinity

therefoooore, (-1.65, .412) is the absolute minimum and the point closest to the origin