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Step-by-Step Solution
Step 1: Understand the Problem
We need to maximize the function
z = 6xy + y^2
subject to the constraints
3x + 4y \le 100,\; 4x + 3y \le 75,\; x \ge 0,\; y \ge 0.
Step 2: Express x in Terms of y
From the constraints:
3x + 4y \le 100 \implies x \le \frac{100 - 4y}{3}
4x + 3y \le 75 \implies x \le \frac{75 - 3y}{4}
Because we want to maximize z = 6xy + y^2 and z increases with larger x (for y \ge 0 ), the best choice for x for any fixed y is the minimum of those upper bounds:
x = \min\left(\frac{100 - 4y}{3}, \frac{75 - 3y}{4}\right).
Step 3: Determine the Feasible Region for y
The constraints x \ge 0 and y \ge 0 also imply:
\frac{100 - 4y}{3} \ge 0 \implies 100 - 4y \ge 0 \implies y \le 25.
\frac{75 - 3y}{4} \ge 0 \implies 75 - 3y \ge 0 \implies y \le 25.
This shows 0 \le y \le 25. Next, compare the two expressions for x over 0 \le y \le 25.
At y = 25, both expressions for x become 0.
When 0 \le y < 25, the value
\frac{75 - 3y}{4}
is typically smaller than
\frac{100 - 4y}{3}
for 0 \le y < 25.
Hence the feasible maximum x over that range is
x = \frac{75 - 3y}{4}
when 0 \le y \le 25.
Step 4: Rewrite the Objective Function in Terms of y Only
Substitute
x = \frac{75 - 3y}{4}
into
z = 6xy + y^2:
z(y) = 6 \left(\frac{75 - 3y}{4}\right) y + y^2.
Expand and simplify:
z(y) = \frac{6y(75 - 3y)}{4} + y^2
= \frac{450y - 18y^2}{4} + y^2
= 112.5y - 4.5y^2 + y^2
= 112.5y - 3.5y^2.
Step 5: Find the Critical Point
To maximize
z(y) = 112.5y - 3.5y^2,
compute the derivative and set it to zero:
\frac{dz}{dy} = 112.5 - 7y = 0
\implies 7y = 112.5
\implies y = \frac{112.5}{7} = \frac{225}{14} \approx 16.07.
This y -value lies in the feasible range [0,25]. Therefore, it is a valid critical point.
Step 6: Determine the Corresponding x and Evaluate z
At y = \frac{225}{14}, the maximum feasible x is
x = \frac{75 - 3y}{4}.
Substitute y = \frac{225}{14} :
x
= \frac{75 - 3\left(\frac{225}{14}\right)}{4}
= \frac{75 - \frac{675}{14}}{4}
= \frac{\frac{1050}{14} - \frac{675}{14}}{4}
= \frac{\frac{375}{14}}{4}
= \frac{375}{56}
\approx 6.696.
Then the maximum value of z is
z\!\Bigl(\frac{225}{14}\Bigr)
= 6 \left(\frac{375}{56}\right)\left(\frac{225}{14}\right) + \left(\frac{225}{14}\right)^2,
which simplifies to approximately 904.02. In exact fractional form, it is
\frac{50625}{56} \approx 904.02.
Hence, the maximum possible value of z is close to 904.
Step 7: Final Answer
The maximum value of z under the given constraints is approximately 904.
Reference Image
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