reza is helping en shah to make a box without the top. the box is made by cutting away four squares from the corners of a 30cm square piece of cardbboard as shown in Figure 1 and bending up the resulting cardboard to form the walls of the box. find the largest possible volume of the box
Let the squares that are cut out each be x by x cm 0<x<15
Now the volume of the prism will be
$$\\V=(30-2x)*(30-2x)*x\\\\
V=x(900-120x+4x^2)\\\\
V=4x^3-120x^2+900x\\\\
\frac{dV}{dx}=12x^2-240x+900\\\\
\frac{d^2V}{dx^2}=24x-240\\\\
$Stationary points will occur when $\frac{dV}{dx}=0\\\
12x^2-240x+900=0\\\\
x^2-20x+75=0\\\\
(x-5)(x-15)=0\\\\
x=5\qquad or \qquad x=15\\\\$$
$$\\When \;\;x=5\;\;\\\\
\frac{d^2V}{dx^2}=24*5-240<0\qquad Maximum \;\;at\;\; x=5\\\\
When \;\;x=15\;\;\\\\
$Volume is 0$\\\\
$So maximum volume occurs when x=5$\\\\
$Maximum volume $= V=(30-10)*(30-10)*5=2000cm^3\\\\$$
Let the squares that are cut out each be x by x cm 0<x<15
Now the volume of the prism will be
$$\\V=(30-2x)*(30-2x)*x\\\\
V=x(900-120x+4x^2)\\\\
V=4x^3-120x^2+900x\\\\
\frac{dV}{dx}=12x^2-240x+900\\\\
\frac{d^2V}{dx^2}=24x-240\\\\
$Stationary points will occur when $\frac{dV}{dx}=0\\\
12x^2-240x+900=0\\\\
x^2-20x+75=0\\\\
(x-5)(x-15)=0\\\\
x=5\qquad or \qquad x=15\\\\$$
$$\\When \;\;x=5\;\;\\\\
\frac{d^2V}{dx^2}=24*5-240<0\qquad Maximum \;\;at\;\; x=5\\\\
When \;\;x=15\;\;\\\\
$Volume is 0$\\\\
$So maximum volume occurs when x=5$\\\\
$Maximum volume $= V=(30-10)*(30-10)*5=2000cm^3\\\\$$