Answer:
Part a) [tex]\frac{1}{2}(4)(2x-3)\leq 10\ in^2[/tex]
Part b) The solution of the inequality Part a) is [tex]x\leq 4[/tex] (see the explanation)
Part c) The maximum height of triangle is 5 inches
Step-by-step explanation:
see the attached figure to better understand the problem
Part a) write an inequality that can be used to find x
we know that
The area of triangle is equal to
[tex]A=\frac{1}{2}bh[/tex]
we have
[tex]b=4\ in[/tex]
[tex]h=(2x-3)\ in[/tex]
[tex]A\leq 10\ in^2[/tex]
so
substitute
[tex]\frac{1}{2}(4)(2x-3)\leq 10\ in^2[/tex] ----> inequality that can can be used to find x
Part b) solve the inequality from part a
we have
[tex]\frac{1}{2}(4)(2x-3)\leq 10[/tex]
solve for x
Simplify left side
[tex]2(2x-3)\leq 10[/tex]
Distribute left side
[tex]4x-6\leq 10[/tex]
Adds 6 both sides
[tex]4x\leq 10+6[/tex]
[tex]4x\leq 16[/tex]
Divide by 4 both sides
[tex]x\leq 16/4[/tex]
[tex]x\leq 4[/tex] ----> solution inequality Part a)
Remember that the height cannot be negative
[tex]h=(2x-3)\ in[/tex]
[tex]h>0[/tex]
[tex](2x-3)> 0[/tex]
[tex]2x > 3[/tex]
[tex]x > 1.5\ in[/tex]
therefore
The value of x must be greater than 1.5 in and less than or equal to 4 in
Part c) what is the maximum height of the triangle?
we know that
The maximum height of triangle will be for the maximum value of x
The maximum value of x is 4 in
[tex]h=(2x-3)\ in[/tex]
substitute for x=4 in
[tex]h=(2(4)-3)=5\ in[/tex]
therefore
The maximum height of triangle is 5 inches
