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)
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)
Let the sides of square base be x and height be h.
volume $\displaystyle = x^2h$
$\displaystyle x^2h=13500$
$\displaystyle h = \frac{13500}{x^2}$
Now, surface area, $\displaystyle A = x^2+4xh$
$\displaystyle A = x^2+4x\left(\frac{13500}{x^2}\right)$
$\displaystyle A = x^2+\frac{54000}{x}$
now, minimize this area, can you take it from here?
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)
$\displaystyle y - 2 = \frac{1}{6}(x+5)$
$\displaystyle -x + 6y = 17$..............(2)
now, solving these two eqns, (1) and (2)
we get (x, y) = (1, 3)
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
$\displaystyle A = x^2+\frac{54000}{x}$
make a table of x and A, taking x = 25 to 32
the minimum A will be when x = 30.
so, $\displaystyle h = \frac{13500}{x^2}=\frac{13500}{30^2}=15$
so, dimensions of box are 30 cm, 30 cm and 15 cm
did you get it now???