(1) Solve the following system of linear equations for the variables x and y:
3x - 2y = 7
2x + y = 4
(2) A large ice cube is melting, each edge shrinking by 2 centimeters per minute. Find the rate at which the volume is changing at the moment when the length of each edge is 15 cm. Assume the cube maintains its shape throughout the melting process.
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(1) To solve the system of linear equations:
3x - 2y = 7 ...(Equation 1)
2x + y = 4 ...(Equation 2)
We can use either the substitution method or the elimination method. Let's use the elimination method:
Multiply Equation 2 by 2 to eliminate the y-term:
4x + 2y = 8 ...(Equation 3)
Now add Equation 1 and Equation 3 to eliminate the y-term:
(3x - 2y) + (4x + 2y) = 7 + 8
7x = 15
Divide both sides by 7 to solve for x:
x = 15/7
Substitute the value of x into either Equation 1 or Equation 2 to solve for y. Let's use Equation 2:
2(15/7) + y = 4
30/7 + y = 4
y = 4 - 30/7
y = 28/7 - 30/7
y = -2/7
Therefore, the solution to the system of equations is:
x = 15/7
y = -2/7
(2) Let V represent the volume of the ice cube, and l represent the length of each edge.
The volume V of a cube is given by V = l^3.
The rate at which the volume is changing is the derivative of the volume with respect to time (dV/dt).
Given that each edge is shrinking by 2 centimeters per minute (dl/dt = -2), we need to find dV/dt when l = 15 cm.
Using the chain rule, we have:
dV/dt = (dV/dl) * (dl/dt)
Differentiating the volume V = l^3 with respect to l, we get:
dV/dl = 3l^2
Substituting dl/dt = -2 and l = 15 into the equation, we have:
dV/dt = (3(15)^2) * (-2)
dV/dt = 3 * 225 * (-2)
dV/dt = -1350
Therefore, at the moment when each edge is 15 cm long, the rate at which the volume is changing is -1350 cubic centimeters per minute.