Rounded to the nearest whole number, the estimated value of the machine after 5 years is $1,566,667.
Given that the rate of depreciation
of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased, we have the relation:
![\[ (dv)/(dt) = k \cdot \frac{1}{{(t+1)}^2} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/6akfk7i3yp524nfovdru3xybw12c5f7wf7.png)
Where:
- V is the value of the machine at time t
- t is the time in years after the machine was purchased
- k is a constant of proportionality
Given that the initial value of the machine was $900,000 and it decreased $200,000 in the first year, we can use this information to estimate the value of k.
Let's set up the initial conditions:
-
(initial value)
-
(value after the first year when t = 1
Using this information, we can solve for k by substituting these values into the equation:
![\[ (dv)/(dt) = k \cdot \frac{1}{{(t+1)}^2} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/6akfk7i3yp524nfovdru3xybw12c5f7wf7.png)
At t = 1:
![\[ (dv)/(dt) = k \cdot \frac{1}{{(1+1)}^2} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/7wtlkojgu7zrv4ds1u3xeg46bp4rdagj2n.png)
![\[ (dv)/(dt) = k \cdot (1)/(4) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/bnbpfdxa3rqoih6024rnolcld52bxrlq6g.png)
Given that
(decrease in value in the first year), we can solve for k:
![\[ \$200,000 = k \cdot (1)/(4) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/g1s5x9p1qelzakjf2z8j1ezxssbse4662k.png)
![\[ k = \$200,000 * 4 = \$800,000 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/h4kqr5l4tjlvrslqqlc4pgtgndi7hxgx2t.png)
Now, with the value of k, we can proceed to estimate the value of the machine after 5 years. We'll integrate the equation:
![\[ (dv)/(dt) = \$800,000 \cdot \frac{1}{{(t+1)}^2} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/mfvmf50fcfacg5ibfxlxbe2fcewq35476b.png)
Integrating this gives:
![\[ V = -\$800,000 \cdot \frac{1}{{t+1}} + C \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/hwl40dcgseqokjvvd24gyb3n4y0oigto7f.png)
Given the initial value of the machine
, substitute t = 0 and solve for C:
![\[ \$900,000 = -\$800,000 \cdot \frac{1}{{0+1}} + C \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/8eel6dhnshdeu8iw98o793549fhovf9ysz.png)
![\[ \$900,000 = -\$800,000 + C \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/vqzwpd0kvt51oqf4op6mv9na1tzawr2plg.png)
![\[ C = \$1,700,000 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/p5qwmsitiz8isjrqtyd9lerbe2hs6e4jn0.png)
Now, for t = 5:
![\[ V = -\$800,000 \cdot \frac{1}{{5+1}} + \$1,700,000 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/7z8n5bv6sso3unibgw0nlhoknc4r3hl8kw.png)
![\[ V = -\$800,000 \cdot (1)/(6) + \$1,700,000 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ly5f2ok571zj2kqf593qhtgrm0fgta8cnv.png)
![\[ V = -\$133,333.33 + \$1,700,000 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/mews325jb8p6thxrbqesf8785c7ichhvcr.png)
![\[ V \approx \$1,566,666.67 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/bt9y6m85c348o45c4ldu3scqzg6vhqm094.png)