Question

A rectangular sheet of fixed perimeter with sides having length in the ratio 8:15 is converted into an open rectangular box by folding after removing squares of equal area

Answer 1

let side of the rectangle be 15k and 8k. and that of square be x.

total are removed = 100

⇒$4x=100$

⇒$x=5$

side of the sqaure will be height of box.

volume of box will be

$(14k-2x)(8k-2x)(x)$

$V=2(2x^{3}-23kx^{2}+60k^{2}x )$

for maximum volume

$(\frac{dV}{dx})_{x=5}   = 0$

$6x^{2} -46kx+60k^{2} =0$

$6(5^{2} )-46k(5)+60k^{2}$

$6k^{2}-23k+15=0$

solving the quadrating eqn we get

$k= 3, \frac{5}{6}$

second derivative of volume is

$\frac{d^{2}V}{dx^{2}}=2(12x-46k)=24x-92k$

and at x=5 and k=5/6

$\frac{d^{2}V}{dx^{2}}=43.34>0$

hence 5/6 is rejected

and at k=3

$\frac{d^{2}V}{dx^{2}}=-201<0$

since the second derivative is less than 0 at k=3...so the volume will be max and hence finally we get x=5 and k=3.

which implies the sides are 15k = 15(3) =45 units and 8k = 8(3) = 24units.

thanks.