Answer:
(a) Maximum slope
,
![H(x)=\frac {1}{12}x](https://img.qammunity.org/2021/formulas/mathematics/college/ncmq6093p322ckfrirny4vyxjj190na1yg.png)
(b) 12.5 in.
Explanation:
The given condition is the wheelchair ramp must have a maximum rise of 1 in. for every horizontal distance of 12 in.
(a) Let m be the maximum allowable slope for the ramp.
![m=\frac{\text{Maximum rise for the given horozontal disence}}{\text{The given horizontal distance}}](https://img.qammunity.org/2021/formulas/mathematics/college/9cj1r8s3b116mp00ltwcb4jwvqonaspxsw.png)
.
Let x be the distance in the horizontal direction as shown in the figure.
So, for slope m, the linear function H for the height is
, where C is constant.
Now, at the starting of the ramp, x=0, and H=0.
Putting this condition back to the equation, we have
![0=m* 0 +C](https://img.qammunity.org/2021/formulas/mathematics/college/mxwe7wbqma9mftp8b47liskgm2e895377m.png)
.
Hence, the required equation is
![H(x)=\frac {1}{12}x](https://img.qammunity.org/2021/formulas/mathematics/college/ncmq6093p322ckfrirny4vyxjj190na1yg.png)
(b) The ramp is 150 in. wide,
So, height gained by the ramp at the end is,
in.