Answer
$dT_L = -40K$
Work Step by Step
From the question, we know that Carnot engine efficiency : $30 \%$ or $0.3$ at $T_H = 400K$
To find the $T_L$ that increases efficiency up to $0.4$, we must first find the difference in efficiency
$\Delta \epsilon = \epsilon_L - \epsilon_H$
$\Delta \epsilon = 0.4 - 0,3 $
$\Delta \epsilon = 0.1 $
Now we go back to the efficiency equation
$\epsilon = 1 - \frac{T_L}{T_H}$
Now differentiate with respect to $T_L$
$d\epsilon = dT_L - \frac{dT_L}{T_H}$
$d\epsilon = dT_L( - \frac{1}{T_H})$
$\frac{d\epsilon}{dT_L} = - \frac{1}{T_H}$
Taking $\Delta \epsilon = 0.1 $ and $T_H = 400K$,
$\frac{0.1}{dT_L} = - \frac{1}{400K}$
$dT_L = -40K$
The temperature of the low-temperature reservoir must be changed to $-40$ to increase the efficiency to 40.0%
