**(a) Maximum thickness of plate:**
Based on puncher strength:

$P = \sigma A$
$P = 50 \, [ \, \frac{1}{4} \pi (2.5^2) \, ]$

$P = 78.125 \pi \, \text{kips}$ → *Equivalent shear force of the plate*

Based on shear strength of plate:

$V = \tau A$ → $V = P$

$78.125\pi = 40 \, [ \, \pi(2.5t) \, ]$

$t = 0.781 \, \text{inch}$ *answer*

**(b) Diameter of smallest hole:**

Based on compression of puncher:

$P = \sigma A$
$P = 50(\frac{1}{4} \pi d^2)$

$P = 12.5 \pi d^2$ → *Equivalent shear force for plate*

Based on shearing of plate:

$V = \tau A$ → $V = P$

$12.5 \pi d^2 = 40 [ \, \pi d(0.25) \, ]$

$d = 0.8 \, \text{in}$ *answer*