# Math Help - optimazation problems

1. ## optimazation problems

A box with a square base and open top must have a volume of 13500cm^3.find the dimension of the box that minize the amount of material used.

2. find the point on the line 6x + y = 9 that is closest to the point (-5,2)

2. Originally Posted by mathsleaner
A box with a square base and open top must have a volume of 13500cm^3.find the dimension of the box that minize the amount of material used.
Let the sides of square base be x and height be h.

volume $= x^2h$

$x^2h=13500$

$h = \frac{13500}{x^2}$

Now, surface area, $A = x^2+4xh$

$A = x^2+4x\left(\frac{13500}{x^2}\right)$

$A = x^2+\frac{54000}{x}$

now, minimize this area, can you take it from here?

3. thanks for the help but how do i minimize, that is the problem i have.how do i minimize the area.

4. Originally Posted by mathsleaner

2. find the point on the line 6x + y = 9 that is closest to the point (-5,2)
6x + y = 9 ...........................(1)

Let the point (x, y) be on the line(1), which is closest to point (-5, 2)

so, A line drawn from (-5, 2) to (x, y) is perpendicular to given line.

so, slope of this line = 1/6

eqn of this line (green dotted line in diagram)

$y - 2 = \frac{1}{6}(x+5)$

$-x + 6y = 17$..............(2)

now, solving these two eqns, (1) and (2)

we get (x, y) = (1, 3)

5. Originally Posted by mathsleaner
thanks for the help but how do i minimize, that is the problem i have.how do i minimize the area.
Let me know which grade are you in? do you know calculus to minimize ???
If not, then make a table of values of x and A
calculate different values of A by taking different x values, by putting in above expression

$A = x^2+\frac{54000}{x}$

make a table of x and A, taking x = 25 to 32

the minimum A will be when x = 30.

so, $h = \frac{13500}{x^2}=\frac{13500}{30^2}=15$

so, dimensions of box are 30 cm, 30 cm and 15 cm

did you get it now???