Math Snap

STEP 1

What is this asking?
How much more money does it cost to make 8 gizmos than to make 4 gizmos, if we know how much the cost changes at each step?
Watch out!
We're given the rate of change of the cost, not the cost itself!

STEP 2

1. Find the cost function
2. Calculate the cost difference

STEP 3

We're given the marginal cost $C^{\prime}(x) = \frac{x^2}{2} - 2x + 29$, which tells us how much the cost changes as we make more gizmos.
To find the total cost of making $x$ gizmos, we need to find a function $C(x)$ whose derivative is $C^{\prime}(x)$.
That means we need to integrate the marginal cost function!

STEP 4

Let's integrate $C^{\prime}(x)$ with respect to $x$:
$$C(x) = \int C^{\prime}(x) \, dx = \int \left( \frac{x^2}{2} - 2x + 29 \right) dx$$ $$C(x) = \frac{x^3}{2 \cdot 3} - \frac{2x^2}{2} + 29x + K = \frac{x^3}{6} - x^2 + 29x + K$$where $K$ is the constant of integration.
This constant represents the fixed costs, like renting the factory, which don't change based on how many gizmos we make.

STEP 5

We want to find the increase in cost when production goes from 4 units to 8 units.
This is the difference between the cost of making 8 units and the cost of making 4 units.
We can write this as $C(8) - C(4)$.

STEP 6

Let's plug in $x = 8$ into our cost function:
$$C(8) = \frac{\textbf{8}^3}{6} - \textbf{8}^2 + 29 \cdot \textbf{8} + K = \frac{512}{6} - 64 + 232 + K = \frac{256}{3} + 168 + K$$

STEP 7

Now let's plug in $x = 4$:
$$C(4) = \frac{\textbf{4}^3}{6} - \textbf{4}^2 + 29 \cdot \textbf{4} + K = \frac{64}{6} - 16 + 116 + K = \frac{32}{3} + 100 + K$$

STEP 8

Finally, let's find the difference:
$$C(8) - C(4) = \left( \frac{256}{3} + 168 + K \right) - \left( \frac{32}{3} + 100 + K \right)$$ $$C(8) - C(4) = \frac{256 - 32}{3} + 168 - 100 + K - K = \frac{224}{3} + 68 = \frac{224 + 204}{3} = \frac{428}{3} \approx 142.67$$

The increase in cost is approximately $\$142.67$.