# Thread: Help with optimization problem?

1. ## Help with optimization problem?

A straight line with a negative slope passes (4,5). Determine the line slope such that the area (in the first quadrant) between this line, x and y axis has a minimum.

I'm totally stuck with this question and don't where to start from. anybody know how to solve this?

2. Let f be the function describing the line. Since it passes through (4,5), $\displaystyle f(x) = 5 - b(x-4)$ where $\displaystyle -b$ (b>0) is the slope of the line. The shape formed in the first quadrant by this line and the axes is a triangle whose height is the y-intercept of the line and whose width is its x-intercept. You can compute the intercepts yourself to verify that they are $\displaystyle \frac5b+4$ and $\displaystyle 5+4b$. The area of the triangle is therefore $\displaystyle \frac12(\frac5b+4)(5+4b) = 8b + \frac{25}{2b} + 20$. Can you take it from here by setting the derivative to be zero? You should obtain $\displaystyle b = 5/4$, I think.

3. Hi thanks for the assistance. I managed to find the height, but for my width i got X = -(5/b) - 4 instead.

Using Y = bX + C, and taking y=0 and C = 5 +4m to find the x intercept,

0=bX + 5 + 4b

X=-(5/b) - 4

May i know where i've gone wrong?

4. Originally Posted by toffeefan
Hi thanks for the assistance. I managed to find the height, but for my width i got X = -(5/b) - 4 instead.

Using Y = bX + C, and taking y=0 and C = 5 +4m to find the x intercept,

0=bX + 5 + 4b

X=-(5/b) - 4

May i know where i've gone wrong?
First you are confusing "m" and "b". But mainly you have the wrong value for C. If Y= bX+ C and X= 4 when Y= 5, 5= 4b+ C so C= 5- 4C, not 5+ 4m or 5+ 4C.